http://vjudge.net/contest/view.action?cid=51211#overview

花了好长时间了，终于把这个专题做了绝大部分了

A：HDU 3853

最简单的概率DP求期望，从终点推到起点就是了，注意一个坑就是如果p1=1那么他一旦到达这个点，那么就永远走不出去了

题解

  1.  //#pragma comment(linker,"/STACK:102400000,102400000")
  2.  #include <map>
  3.  #include <set>
  4.  #include <stack>
  5.  #include <queue>
  6.  #include <cmath>
  7.  #include <ctime>
  8.  #include <vector>
  9.  #include <cstdio>
  10.  #include <cctype>
  11.  #include <cstring>
  12.  #include <cstdlib>
  13.  #include <iostream>
  14.  #include <algorithm>
  15.  using namespace std;
  16.  #define INF 1e9
  17.  #define inf (-((LL)1<<40))
  18.  #define lson k<<1, L, mid
  19.  #define rson k<<1|1, mid+1, R
  20.  #define mem0(a) memset(a,0,sizeof(a))
  21.  #define mem1(a) memset(a,-1,sizeof(a))
  22.  #define mem(a, b) memset(a, b, sizeof(a))
  23.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  24.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  25.  
  26.  //typedef __int64 LL;
  27.  //typedef long long LL;
  28.  const int MAXN = ;
  29.  const int MAXM = ;
  30.  const double eps = 1e-;
  31.  //const LL MOD = 1000000007;
  32.  
  33.  double p1[MAXN][MAXN], p2[MAXN][MAXN], p3[MAXN][MAXN], dp[MAXN][MAXN];
  34.  
  35.  int main()
  36.  {
  37.      int R, C;
  38.      while(~scanf("%d %d", &R, &C))
  39.      {
  40.          for(int i=;i<=R;i++)
  41.              for(int j=;j<=C;j++)
  42.                  scanf("%lf%lf%lf", &p1[i][j], &p2[i][j], &p3[i][j]);
  43.          mem0(dp);
  44.          for(int i=R;i>=;i--)
  45.              for(int j=C;j>=;j--)
  46.              {
  47.                  if(i==R && j==C) continue;
  48.                  if(fabs(p1[i][j] - ) < eps) continue;
  49.                  dp[i][j] = (dp[i][j+]*p2[i][j] + dp[i+][j]*p3[i][j] + ) / (-p1[i][j]) ;
  50.              }
  51.          printf("%.3lf\n", dp[][]);
  52.      }
  53.      return ;
  54.  }

B:HDU 3920

状态压缩DP  题解

  1.  //#pragma comment(linker,"/STACK:102400000,102400000")
  2.  #include <map>
  3.  #include <set>
  4.  #include <stack>
  5.  #include <queue>
  6.  #include <cmath>
  7.  #include <ctime>
  8.  #include <vector>
  9.  #include <cstdio>
  10.  #include <cctype>
  11.  #include <cstring>
  12.  #include <cstdlib>
  13.  #include <iostream>
  14.  #include <algorithm>
  15.  using namespace std;
  16.  #define INF 1e9
  17.  #define inf (-((LL)1<<40))
  18.  #define lson k<<1, L, mid
  19.  #define rson k<<1|1, mid+1, R
  20.  #define mem0(a) memset(a,0,sizeof(a))
  21.  #define mem1(a) memset(a,-1,sizeof(a))
  22.  #define mem(a, b) memset(a, b, sizeof(a))
  23.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  24.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  25.  template<class T> T CMP_MIN(T a, T b) { return a < b; }
  26.  template<class T> T CMP_MAX(T a, T b) { return a > b; }
  27.  template<class T> T MAX(T a, T b) { return a > b ? a : b; }
  28.  template<class T> T MIN(T a, T b) { return a < b ? a : b; }
  29.  template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
  30.  template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
  31.  
  32.  //typedef __int64 LL;
  33.  //typedef long long LL;
  34.  const int MAXN = ;
  35.  const int MAXM = ;
  36.  const double eps = 1e-;
  37.  //const LL MOD = 1000000007;
  38.  
  39.  int T, N;
  40.  typedef double Point[];
  41.  Point st, p[MAXN];
  42.  double dis[MAXN][MAXN], d[MAXN], dp[<<];
  43.  
  44.  double calc(Point a, Point b)
  45.  {
  46.      double x = a[] - b[];
  47.      double y = a[] - b[];
  48.      return sqrt(x*x + y*y);
  49.  }
  50.  
  51.  void getDis()
  52.  {
  53.      for(int i=;i<N;i++)
  54.      {
  55.          d[i] = calc(st, p[i]);
  56.          for(int j=i+;j<N;j++)
  57.          {
  58.              dis[j][i] = dis[i][j] = calc(p[i], p[j]);
  59.          }
  60.      }
  61.  }
  62.  
  63.  int main()
  64.  {
  65.      int t = ;
  66.      scanf("%d", &T);
  67.      while(T--)
  68.      {
  69.          scanf("%lf %lf", &st[], &st[]);
  70.          scanf("%d", &N);
  71.          N <<= ;
  72.          for(int i=;i<N;i++)
  73.              scanf("%lf %lf", &p[i][], &p[i][]);
  74.          getDis();
  75.          dp[] = ;
  76.          for(int i=;i<(<<N);i++) dp[i] = INF;
  77.          queue<int>q;
  78.          q.push();
  79.          while(!q.empty())
  80.          {
  81.              int now = q.front(); q.pop();
  82.              int f=, r;
  83.              while( now & (<<f)  && f < N)
  84.                  f++;
  85.              for(r = f + ; r < N; r ++ )
  86.                  if(!(now & (<<r)))
  87.              {
  88.                  int next = now | (<<f) | (<<r);
  89.                  double minDis = MIN(d[f], d[r]) + dis[f][r];
  90.                  if( fabs(dp[next] - INF) < eps )
  91.                  {
  92.                      q.push(next);
  93.                      dp[next] = dp[now] + minDis;
  94.                  }
  95.                  else if( dp[now] + minDis < dp[next] )
  96.                      dp[next] = dp[now] + minDis;
  97.              }
  98.          }
  99.          printf("Case #%d: %.2lf%\n", ++t, dp[(<<N)-]);
  100.      }
  101.      return ;
  102.  }

C:HDU 1520

简单的树形DP  题解

  1.  //#pragma comment(linker,"/STACK:102400000,102400000")
  2.  #include <map>
  3.  #include <set>
  4.  #include <stack>
  5.  #include <queue>
  6.  #include <cmath>
  7.  #include <ctime>
  8.  #include <vector>
  9.  #include <cstdio>
  10.  #include <cctype>
  11.  #include <cstring>
  12.  #include <cstdlib>
  13.  #include <iostream>
  14.  #include <algorithm>
  15.  using namespace std;
  16.  #define INF 1e9
  17.  #define inf (-((LL)1<<40))
  18.  #define lson k<<1, L, mid
  19.  #define rson k<<1|1, mid+1, R
  20.  #define mem0(a) memset(a,0,sizeof(a))
  21.  #define mem1(a) memset(a,-1,sizeof(a))
  22.  #define mem(a, b) memset(a, b, sizeof(a))
  23.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  24.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  25.  template<class T> T CMP_MIN(T a, T b) { return a < b; }
  26.  template<class T> T CMP_MAX(T a, T b) { return a > b; }
  27.  template<class T> T MAX(T a, T b) { return a > b ? a : b; }
  28.  template<class T> T MIN(T a, T b) { return a < b ? a : b; }
  29.  template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
  30.  template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
  31.  
  32.  //typedef __int64 LL;
  33.  //typedef long long LL;
  34.  const int MAXN = ;
  35.  const int MAXM = ;
  36.  const double eps = 1e-;
  37.  //const LL MOD = 1000000007;
  38.  
  39.  int N, a[MAXN], dp[MAXN][];
  40.  int fa[MAXN];
  41.  vector<int>e[MAXN];
  42.  
  43.  void DFS(int u)
  44.  {
  45.      int s0 = , s1 = ;
  46.      for(int i=;i<e[u].size();i++)
  47.      {
  48.          DFS(e[u][i]);
  49.          s0 += MAX( dp[e[u][i]][], dp[e[u][i]][] );
  50.          s1 += dp[e[u][i]][];
  51.      }
  52.      dp[u][] = s0;
  53.      dp[u][] = s1 + a[u];
  54.  }
  55.  
  56.  int main()
  57.  {
  58.      //FOPENIN("in.txt");
  59.      while(~scanf("%d", &N))
  60.      {
  61.          mem0(dp);
  62.          for(int i=;i<=N;i++)
  63.          {
  64.              scanf("%d", &a[i]);
  65.              fa[i] = i;
  66.              e[i].clear();
  67.          }
  68.          int x, y;
  69.          while(scanf("%d %d", &x, &y) && (x||y) ){
  70.              e[y].push_back(x);
  71.              fa[x] = y;
  72.          }
  73.          int ans = ;
  74.          for(int i=;i<=N;i++) if(fa[i] == i)
  75.          {
  76.              DFS(i);
  77.              ans += MAX(dp[i][], dp[i][]);
  78.          }
  79.          printf("%d\n", ans);
  80.      }
  81.      return ;
  82.  }

D:HDU 4284

状态压缩DP  题解

  1.  //#pragma comment(linker,"/STACK:102400000,102400000")
  2.  #include <map>
  3.  #include <set>
  4.  #include <stack>
  5.  #include <queue>
  6.  #include <cmath>
  7.  #include <ctime>
  8.  #include <vector>
  9.  #include <cstdio>
  10.  #include <cctype>
  11.  #include <cstring>
  12.  #include <cstdlib>
  13.  #include <iostream>
  14.  #include <algorithm>
  15.  using namespace std;
  16.  #define INF 1e8
  17.  #define inf (-((LL)1<<40))
  18.  #define lson k<<1, L, mid
  19.  #define rson k<<1|1, mid+1, R
  20.  #define mem0(a) memset(a,0,sizeof(a))
  21.  #define mem1(a) memset(a,-1,sizeof(a))
  22.  #define mem(a, b) memset(a, b, sizeof(a))
  23.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  24.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  25.  template<class T> T CMP_MIN(T a, T b) { return a < b; }
  26.  template<class T> T CMP_MAX(T a, T b) { return a > b; }
  27.  template<class T> T MAX(T a, T b) { return a > b ? a : b; }
  28.  template<class T> T MIN(T a, T b) { return a < b ? a : b; }
  29.  template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
  30.  template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
  31.  
  32.  //typedef __int64 LL;
  33.  //typedef long long LL;
  34.  const int MAXN = ;
  35.  const int MAXM = ;
  36.  const double eps = 1e-;
  37.  //const LL MOD = 1000000007;
  38.  
  39.  int T, N, M, H, Money;
  40.  int dis[MAXN][MAXN], add[MAXN], cost[MAXN];
  41.  int citys[], dp[<<][];
  42.  
  43.  void init()
  44.  {
  45.      int u, v, w, c, a;
  46.       mem0(add); mem0(cost);mem1(dp);
  47.      scanf("%d %d %d", &N, &M, &Money);
  48.      for(int i = ; i <= N ;i ++ )
  49.      {
  50.          dis[i][i] = ;
  51.          for(int j = ; j <= N ; j ++ )
  52.              if(i != j) dis[i][j] = INF;
  53.      }
  54.      for(int i = ; i < M; i ++ )
  55.      {
  56.          scanf("%d %d %d", &u, &v, &w);
  57.          dis[u][v] = dis[v][u] = MIN(dis[u][v], w);
  58.      }
  59.      scanf("%d", &H);
  60.      for(int i = ; i <= H; i ++ )
  61.      {
  62.          scanf("%d %d %d", &u, &a, &c);
  63.          citys[i] = u;
  64.          add[i] = a;
  65.          cost[i] = c;
  66.      }
  67.  }
  68.  
  69.  void floyd()
  70.  {
  71.      for(int k=;k<=N;k++)
  72.      for(int i=;i<=N;i++)
  73.      for(int j=;j<=N;j++)
  74.      {
  75.          dis[i][j] = MIN(dis[i][j], dis[i][k] + dis[k][j]);
  76.      }
  77.  }
  78.  
  79.  int main()
  80.  {
  81.      //FOPENIN("in.txt");
  82.      while(~scanf("%d", &T)) while(T--)
  83.      {
  84.          init();
  85.          floyd();
  86.          int ans = -INF; H += ;
  87.          citys[] = ; cost[] = add[] = ;
  88.          dp[][] = Money;
  89.          for(int now = ; now < (<<H); now ++ )
  90.          {
  91.              for(int u = ; u < H; u ++ ) if(dp[now][u] != -)
  92.              {
  93.                  for(int v = ; v < H; v ++ ) if( (now & (<<v)) ==  )
  94.                  {
  95.                      int next = now | (<<v);
  96.                      if(dp[now][u] >= dis[citys[u]][citys[v]] + cost[v] )
  97.                      {
  98.                          dp[next][v] = MAX(dp[now | (<<v)][v], dp[now][u] - dis[citys[u]][citys[v]] - cost[v] + add[v]);
  99.                      }
  100.                      if(next == (<<H) - )
  101.                      {
  102.                          ans = MAX(ans, dp[next][v]);
  103.                      }
  104.                  }
  105.              }
  106.          }
  107.         //printf("%d\n", ans);
  108.          printf("%s\n", ans >=  ? "YES" : "NO");
  109.      }
  110.      return ;
  111.  }

E:HDU 2196

比较好的题目了，两次DFS，求树上任意点触发的最长距离 题解

  1.  //#pragma comment(linker,"/STACK:102400000,102400000")
  2.  #include <map>
  3.  #include <set>
  4.  #include <stack>
  5.  #include <queue>
  6.  #include <cmath>
  7.  #include <ctime>
  8.  #include <vector>
  9.  #include <cstdio>
  10.  #include <cctype>
  11.  #include <cstring>
  12.  #include <cstdlib>
  13.  #include <iostream>
  14.  #include <algorithm>
  15.  using namespace std;
  16.  #define INF 1e8
  17.  #define inf (-((LL)1<<40))
  18.  #define lson k<<1, L, mid
  19.  #define rson k<<1|1, mid+1, R
  20.  #define mem0(a) memset(a,0,sizeof(a))
  21.  #define mem1(a) memset(a,-1,sizeof(a))
  22.  #define mem(a, b) memset(a, b, sizeof(a))
  23.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  24.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  25.  template<class T> T CMP_MIN(T a, T b) { return a < b; }
  26.  template<class T> T CMP_MAX(T a, T b) { return a > b; }
  27.  template<class T> T MAX(T a, T b) { return a > b ? a : b; }
  28.  template<class T> T MIN(T a, T b) { return a < b ? a : b; }
  29.  template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
  30.  template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
  31.  
  32.  //typedef __int64 LL;
  33.  //typedef long long LL;
  34.  const int MAXN = ;
  35.  const int MAXM = ;
  36.  const double eps = 1e-;
  37.  //const LL MOD = 1000000007;
  38.  
  39.  int N;
  40.  int head[MAXN], next[MAXM], tot;
  41.  int  u[MAXM], v[MAXM], w[MAXM];
  42.  int fir[MAXN], sec[MAXN], ans[MAXN];
  43.  
  44.  void addEdge(int U, int V, int W)
  45.  {
  46.      u[tot] = U;
  47.      v[tot] = V;
  48.      w[tot] = W;
  49.      next[tot] = head[U];
  50.      head[U] = tot;
  51.      tot ++;
  52.  }
  53.  
  54.  int dfs1(int x, int fa)
  55.  {
  56.      fir[x] = sec[x] = ;
  57.      for(int e = head[x]; e != -; e = next[e]) if(v[e] != fa)
  58.      {
  59.          int dis = dfs1(v[e], x) + w[e];
  60.          if(dis >= fir[x]) { sec[x] = fir[x]; fir[x] = dis; }
  61.          else if(dis > sec[x]) sec[x] = dis;
  62.      }
  63.      return fir[x];
  64.  }
  65.  
  66.  void dfs2(int x, int fa, int dis)
  67.  {
  68.      ans[x] = MAX(fir[x], dis);
  69.      for(int e = head[x]; e != -; e = next[e]) if(v[e] != fa)
  70.      {
  71.          int y = v[e];
  72.          if(fir[y] + w[e] == fir[x])
  73.              dfs2(y, x, MAX( dis, sec[x]) + w[e] );
  74.          else
  75.              dfs2(y, x, MAX( dis, fir[x]) + w[e] );
  76.      }
  77.  }
  78.  
  79.  int main()
  80.  {
  81.  
  82.      while(~scanf("%d", &N))
  83.      {
  84.          tot = ;
  85.          mem1(head);
  86.          int V, W;
  87.          for(int i = ; i <= N; i ++)
  88.          {
  89.              scanf("%d %d", &V, &W);
  90.              addEdge(i, V, W);
  91.              addEdge(V, i, W);
  92.          }
  93.          dfs1(, );
  94.          dfs2(, , );
  95.          for(int i = ; i <= N; i ++ )
  96.              printf("%d\n", ans[i]);
  97.      }
  98.      return ;
  99.  }

F:HDU 4539

最原始最薄里的状压DP  题解

  1.  //#pragma comment(linker,"/STACK:102400000,102400000")
  2.  #include <map>
  3.  #include <set>
  4.  #include <stack>
  5.  #include <queue>
  6.  #include <cmath>
  7.  #include <ctime>
  8.  #include <vector>
  9.  #include <cstdio>
  10.  #include <cctype>
  11.  #include <cstring>
  12.  #include <cstdlib>
  13.  #include <iostream>
  14.  #include <algorithm>
  15.  using namespace std;
  16.  #define INF 1e8
  17.  #define inf (-((LL)1<<40))
  18.  #define lson k<<1, L, mid
  19.  #define rson k<<1|1, mid+1, R
  20.  #define mem0(a) memset(a,0,sizeof(a))
  21.  #define mem1(a) memset(a,-1,sizeof(a))
  22.  #define mem(a, b) memset(a, b, sizeof(a))
  23.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  24.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  25.  template<class T> T CMP_MIN(T a, T b) { return a < b; }
  26.  template<class T> T CMP_MAX(T a, T b) { return a > b; }
  27.  template<class T> T MAX(T a, T b) { return a > b ? a : b; }
  28.  template<class T> T MIN(T a, T b) { return a < b ? a : b; }
  29.  template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
  30.  template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
  31.  
  32.  //typedef __int64 LL;
  33.  //typedef long long LL;
  34.  const int MAXN = ;
  35.  const int MAXM = ;
  36.  const double eps = 1e-;
  37.  //const LL MOD = 1000000007;
  38.  
  39.  int N, M;
  40.  int ma[];
  41.  int dp[][][];
  42.  
  43.  int stNum;
  44.  int st[], num[];
  45.  
  46.  int judge(int x)
  47.  {
  48.      if(x & (x<<)) return ;
  49.      return ;
  50.  }
  51.  
  52.  int get1(int x)
  53.  {
  54.      int ret = ;
  55.      while(x)
  56.      {
  57.          ret += x & ;
  58.          x >>= ;
  59.      }
  60.      return ret;
  61.  }
  62.  
  63.  void init()
  64.  {
  65.      mem1(dp);mem0(ma);
  66.      stNum = ;
  67.      for(int i=;i<(<<M);i++) if(judge(i))
  68.      {
  69.          st[stNum] = i;
  70.          num[stNum++] = get1(i);
  71.      }
  72.  }
  73.  
  74.  int main()
  75.  {
  76.      //FOPENIN("in.txt");
  77.      while(~scanf("%d %d%*c", &N, &M))
  78.      {
  79.          init();
  80.          int x, cur = ;
  81.          for(int i=;i<=N;i++)
  82.              for(int j=;j<M;j++) {
  83.                  scanf("%d", &x);
  84.                  if(x==) ma[i] |= (<<j);
  85.              }
  86.          for(int i=;i<stNum;i++) {
  87.              if(st[i] & ma[]) continue;
  88.              dp[cur][i][] = num[i];
  89.          }
  90.          for(int r = ; r <= N; r ++)
  91.          {
  92.              cur = !cur;
  93.              for(int i = ; i < stNum; i ++)
  94.              {
  95.                  if(st[i] & ma[r]) continue;
  96.                  for(int j = ; j < stNum; j ++ )
  97.                  {
  98.                      if(st[j] & ma[r-]) continue;
  99.                      int p = (st[i]<<) | (st[i]>>);
  100.                      if(st[j] & p) continue;
  101.                      for(int k = ; k < stNum; k ++) if(dp[!cur][j][k] != -)
  102.                      {
  103.                          int pp = st[i] | ((st[j]<<) | (st[j]>>));
  104.                          if(st[k] & pp) continue;
  105.                          dp[cur][i][j] = MAX(dp[cur][i][j], dp[!cur][j][k] + num[i]);
  106.                      }
  107.  
  108.                  }
  109.              }
  110.          }
  111.          int ans = ;
  112.          for(int i=;i<stNum;i++) for(int j=;j<stNum;j++)
  113.              ans = MAX(ans, dp[cur][i][j]);
  114.          printf("%d\n", ans);
  115.      }
  116.      return ;
  117.  }
  118.  
  119.  /*
  120.  
  121.  6
  122.  3
  123.  2
  124.  12
  125.  
  126.  */

G:HDU 4035

推公式很重要，是看别人做的。。。。题解

  1.  //#pragma comment(linker,"/STACK:102400000,102400000")
  2.  #include <map>
  3.  #include <set>
  4.  #include <stack>
  5.  #include <queue>
  6.  #include <cmath>
  7.  #include <ctime>
  8.  #include <vector>
  9.  #include <cstdio>
  10.  #include <cctype>
  11.  #include <cstring>
  12.  #include <cstdlib>
  13.  #include <iostream>
  14.  #include <algorithm>
  15.  using namespace std;
  16.  #define INF 1e8
  17.  #define inf (-((LL)1<<40))
  18.  #define lson k<<1, L, mid
  19.  #define rson k<<1|1, mid+1, R
  20.  #define mem0(a) memset(a,0,sizeof(a))
  21.  #define mem1(a) memset(a,-1,sizeof(a))
  22.  #define mem(a, b) memset(a, b, sizeof(a))
  23.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  24.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  25.  template<class T> T CMP_MIN(T a, T b) { return a < b; }
  26.  template<class T> T CMP_MAX(T a, T b) { return a > b; }
  27.  template<class T> T MAX(T a, T b) { return a > b ? a : b; }
  28.  template<class T> T MIN(T a, T b) { return a < b ? a : b; }
  29.  template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
  30.  template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
  31.  
  32.  //typedef __int64 LL;
  33.  //typedef long long LL;
  34.  const int MAXN = ;
  35.  const int MAXM = ;
  36.  const double eps = 1e-;
  37.  //const LL MOD = 1000000007;
  38.  
  39.  int T, N;
  40.  vector<int>v[MAXN];
  41.  double k[MAXN], e[MAXN];
  42.  double A[MAXN], B[MAXN], C[MAXN];
  43.  
  44.  void init()
  45.  {
  46.      int U, V;
  47.      scanf("%d", &N);
  48.      for(int i=;i<=N;i++) v[i].clear();
  49.      for(int i=;i<N-;i++)
  50.      {
  51.          scanf("%d %d", &U, &V);
  52.          v[U].push_back(V);
  53.          v[V].push_back(U);
  54.      }
  55.      for(int i=;i<=N;i++)
  56.      {
  57.          scanf("%d %d", &U, &V);
  58.          k[i] = (double)U / 100.0;
  59.          e[i] = (double)V / 100.0;
  60.      }
  61.  }
  62.  
  63.  bool DFS(int x, int fa)
  64.  {
  65.      A[x] = k[x];
  66.      B[x] = ( - k[x] - e[x]) / v[x].size();
  67.      C[x] =  - k[x] - e[x];
  68.      if(v[x].size() ==  && x != fa)
  69.          return true;
  70.      double temp = ;
  71.      for(int i = ; i < v[x].size() ; i ++ )
  72.      {
  73.          int y = v[x][i];
  74.          if(y == fa) continue;
  75.          if(!DFS(y, x)) return false;
  76.          A[x] += A[y] * B[x];
  77.          C[x] += C[y] * B[x];
  78.          temp += B[y] * B[x];
  79.      }
  80.      if(fabs(temp - 1.0) < eps) return false;
  81.      A[x] = A[x] / ( - temp);
  82.      B[x] = B[x] / ( - temp);
  83.      C[x] = C[x] / ( - temp);
  84.      return true;
  85.  }
  86.  
  87.  int main()
  88.  {
  89.      //FOPENIN("in.txt");
  90.      scanf("%d", &T);
  91.      for(int t = ; t <= T; t ++ )
  92.      {
  93.          init();
  94.          if( DFS(, ) && fabs(A[] - 1.0) > eps )
  95.          {
  96.              printf("Case %d: %lf\n", t, C[] / ( - A[]));
  97.          }
  98.          else
  99.          {
  100.              printf("Case %d: impossible\n", t);
  101.          }
  102.      }
  103.  }

I:HDU 1561

树形DP，注意状态转化的过程  题解

  1.  #include <map>
  2.  #include <set>
  3.  #include <stack>
  4.  #include <queue>
  5.  #include <cmath>
  6.  #include <ctime>
  7.  #include <vector>
  8.  #include <cstdio>
  9.  #include <cctype>
  10.  #include <cstring>
  11.  #include <cstdlib>
  12.  #include <iostream>
  13.  #include <algorithm>
  14.  using namespace std;
  15.  #define INF ((LL)100000000000000000)
  16.  #define inf (-((LL)1<<40))
  17.  #define lson k<<1, L, mid
  18.  #define rson k<<1|1, mid+1, R
  19.  #define mem0(a) memset(a,0,sizeof(a))
  20.  #define mem1(a) memset(a,-1,sizeof(a))
  21.  #define mem(a, b) memset(a, b, sizeof(a))
  22.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  23.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  24.  template<class T> T CMP_MIN(T a, T b) { return a < b; }
  25.  template<class T> T CMP_MAX(T a, T b) { return a > b; }
  26.  template<class T> T MAX(T a, T b) { return a > b ? a : b; }
  27.  template<class T> T MIN(T a, T b) { return a < b ? a : b; }
  28.  template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
  29.  template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
  30.  
  31.  //typedef __int64 LL;
  32.  typedef long long LL;
  33.  const int MAXN = ;
  34.  const int MAXM = ;
  35.  const double eps = 1e-;
  36.  int dx[] = {-, , , };
  37.  int dy[] = {, -, , };
  38.  
  39.  int N, M;
  40.  int G[MAXN][MAXN], vis[MAXN], num[MAXN];
  41.  int DP[MAXN][MAXN], D[MAXN];
  42.  
  43.  void DFS(int u)
  44.  {
  45.      vis[u] = ;
  46.      DP[u][] = num[u];
  47.      for(int i=;i<=N;i++)  if(G[u][i])
  48.      {
  49.          if(!vis[i]) DFS(i);
  50.          for(int j=M;j>;j--)//这里为了保证是先从父节点更新，需要逆序
  51.              for(int k=;k<j;k++)//k<j保证父节点不会被更新掉
  52.                  DP[u][j] = MAX(DP[u][j], DP[u][j-k] + DP[i][k]);
  53.      }
  54.  }
  55.  
  56.  int main()
  57.  {
  58.      //FOPENIN("in.txt");
  59.      while(~scanf("%d %d", &N, &M) &&( N||M))
  60.      {
  61.          int a;mem0(G);mem0(D);mem0(vis);mem0(DP);
  62.          for(int i=;i<=N;i++) {
  63.              scanf("%d %d", &a, &num[i]);
  64.              if(a != )G[a][i] = ;
  65.              else D[i] = ;
  66.          }
  67.          for(int i=;i<=N;i++) if(!vis[i]){
  68.              DFS(i);
  69.              if(D[i]) for(int j=M;j>=;j--)//同样需要逆序，可以取到0
  70.              {
  71.                  for(int k=;k<=j;k++)
  72.                      DP[][j] = MAX(DP[][j], DP[][j-k] + DP[i][k]);
  73.              }
  74.          }
  75.          printf("%d\n", DP[][M]);
  76.  
  77.      }
  78.      return ;
  79.  }

J:HDU 4870

1。高斯消元的做法

2。公式推倒很重要

  1.  #include <map>
  2.  #include <set>
  3.  #include <stack>
  4.  #include <queue>
  5.  #include <cmath>
  6.  #include <ctime>
  7.  #include <vector>
  8.  #include <cstdio>
  9.  #include <cctype>
  10.  #include <cstring>
  11.  #include <cstdlib>
  12.  #include <iostream>
  13.  #include <algorithm>
  14.  using namespace std;
  15.  #define INF ((LL)100000000000000000)
  16.  #define inf (-((LL)1<<40))
  17.  #define lson k<<1, L, mid
  18.  #define rson k<<1|1, mid+1, R
  19.  #define mem0(a) memset(a,0,sizeof(a))
  20.  #define mem1(a) memset(a,-1,sizeof(a))
  21.  #define mem(a, b) memset(a, b, sizeof(a))
  22.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  23.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  24.  template<class T> T CMP_MIN(T a, T b) { return a < b; }
  25.  template<class T> T CMP_MAX(T a, T b) { return a > b; }
  26.  template<class T> T MAX(T a, T b) { return a > b ? a : b; }
  27.  template<class T> T MIN(T a, T b) { return a < b ? a : b; }
  28.  template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
  29.  template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
  30.  
  31.  //typedef __int64 LL;
  32.  typedef long long LL;
  33.  const int MAXN = ;
  34.  const int MAXM = ;
  35.  const double eps = 1e-;
  36.  
  37.  double t[];
  38.  
  39.  int main()
  40.  {
  41.      //FOPENIN("in.txt");
  42.      double p;
  43.      while(~scanf("%lf", &p))
  44.      {
  45.          t[] = ;
  46.          t[] =  / p;
  47.          t[] =  / p +  / p / p;
  48.          for(int i=;i<=;i++)
  49.          {
  50.              t[i] =  / p * t[i-] +  / p - (-p) / p * t[i-];
  51.          }
  52.          double ans =  *t[] + t[] - t[];
  53.          printf("%lf\n", ans);
  54.      }
  55.      return ;
  56.  }

K:POJ 1160

我就喜欢暴力求解  题解

  1.  #include <map>
  2.  #include <set>
  3.  #include <stack>
  4.  #include <queue>
  5.  #include <cmath>
  6.  #include <ctime>
  7.  #include <vector>
  8.  #include <cstdio>
  9.  #include <cctype>
  10.  #include <cstring>
  11.  #include <cstdlib>
  12.  #include <iostream>
  13.  #include <algorithm>
  14.  using namespace std;
  15.  #define INF 100000000
  16.  #define inf (-((LL)1<<40))
  17.  #define lson k<<1, L, mid
  18.  #define rson k<<1|1, mid+1, R
  19.  #define mem0(a) memset(a,0,sizeof(a))
  20.  #define mem1(a) memset(a,-1,sizeof(a))
  21.  #define mem(a, b) memset(a, b, sizeof(a))
  22.  #define FOPENIN(IN) freopen(IN, "r", stdin)
  23.  #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
  24.  template<class T> T CMP_MIN ( T a, T b ) { return a < b;   }
  25.  template<class T> T CMP_MAX ( T a, T b ) { return a > b;   }
  26.  template<class T> T MAX ( T a, T b ) { return a > b ? a : b; }
  27.  template<class T> T MIN ( T a, T b ) { return a < b ? a : b; }
  28.  template<class T> T GCD ( T a, T b ) { return b ? GCD ( b, a % b ) : a; }
  29.  template<class T> T LCM ( T a, T b ) { return a / GCD ( a, b ) * b;       }
  30.  template<class T> T SWAP( T& a, T& b ) { T t = a; a = b;  b = t; }
  31.  
  32.  //typedef __int64 LL;
  33.  typedef long long LL;
  34.  const int MAXN = ;
  35.  const int MAXM = ;
  36.  const double eps = 1e-;
  37.  
  38.  int V, P;
  39.  int x[], DP[][][];
  40.  int dis[][], pre[];
  41.  
  42.  void initDis()
  43.  {
  44.          mem0(dis);mem0(pre);
  45.          for(int i=;i<=V;i++)
  46.          for(int j=i;j<=V;j++)
  47.          for(int k=i;k<=j;k++)
  48.          dis[i][j] += MIN(x[k]-x[i], x[j]-x[k]);
  49.          for(int i=;i<=V;i++)for(int j=i;j>=;j--)
  50.                  pre[i] += x[i] - x[j];
  51.  }
  52.  
  53.  int main()
  54.  {
  55.          //FOPENIN ( "in.txt" );
  56.         //FOPENOUT("out.txt");
  57.         while(~scanf("%d %d", &V, &P))
  58.         {
  59.                  for(int i=;i<=V;i++)
  60.                          scanf("%d", &x[i]);
  61.                  initDis();
  62.                  for(int i=;i<;i++)
  63.                  for(int j=;j<=MIN(i,P);j++)
  64.                  for(int k=;k<=i;k++){
  65.                          DP[i][j][k] = INF;
  66.                  }
  67.                  int now = ;
  68.                  for(int i=;i<=V;i++)
  69.                  {
  70.                          now = !now;
  71.                          int s = ;
  72.                          for(int j=;j<=i;j++)   DP[now][][j] = pre[j];
  73.                          for(int j=;j<=MIN(i, P); j++)
  74.                          {
  75.                                  DP[now][j][i] = INF;
  76.                                  for(int k=j-;k<i;k++)if(DP[!now][j-][k] != INF)
  77.                                  {
  78.                                          DP[now][j][i] = MIN(DP[now][j][i], DP[now][j-][k] + dis[k][i]);
  79.                                  }
  80.                                  for(int k=j;k<i;k++) DP[now][j][k] = DP[!now][j][k];
  81.                          }
  82.                  }
  83.                  int ans = INF;
  84.                  for(int k=P;k<=V;k++)if(DP[now][P][k] != INF)
  85.                  {
  86.                          int s = ;
  87.                          for(int j = k + ; j <= V; j ++ ) s += x[j] - x[k];
  88.                          ans = MIN(ans, DP[now][P][k] + s);
  89.                  }
  90.                  printf("%d\n", ans);
  91.         }
  92.          return ;
  93.  }

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