HDU 2016.11.12 做题感想

　　细数一下这两天做过的值得总结的一些题Orz......

　　HDU 2571 简单dp，但是一开始WA了一发。原因很简单：没有考虑仔细。

　　如果指向该点的所有点权值都为负数，那就错了（我一开始默认初始值为0）

　　这是非常基础的典型DAG模型，好久不做，手明显生了……

 #include <bits/stdc++.h>
 
 using namespace std;
 
 #define REP(i,n)                for(int i(0); i <  (n); ++i)
 #define rep(i,a,b)              for(int i(a); i <= (b); ++i)
 #define dec(i,a,b)              for(int i(a); i >= (b); --i)
 #define for_edge(i,x)           for(int i = H[x]; i; i = X[i])
 
 #define LL      long long
 #define ULL     unsigned long long
 #define MP      make_pair
 #define PB      push_back
 #define FI      first
 #define SE      second
 #define INF     1 << 30
 
 const int N     =          +       ;
 const int M     =           +       ;
 const int Q     =            +       ;
 const int A     =              +       ;
 
 int f[A][Q], c[A][Q];
 int T;
 int n, m;
 
 int main(){
 #ifndef ONLINE_JUDGE
     freopen("test.txt", "r", stdin);
     freopen("test.out", "w", stdout);
 #endif
 
     scanf("%d ", &T);
     REP(Case, T){
         scanf("%d %d ", &n, &m);
         memset(f, , sizeof f);
         rep(i, , n) rep(j, , m) scanf("%d ", &c[i][j]);
         f[][] = c[][];
         rep(i, , m){
             int now = f[][i - ];
             rep(j, , i >> ) if (i % j == ) now = max(now, f[][j]);
             f[][i] = c[][i] + now;
         }
 
         rep(i, , n) f[i][] = c[i][] + f[i - ][];
         rep(i, , n) rep(j, , m){
             int now = f[i][j - ];
             rep(k, , j >> ) if (j % k == ) now = max(now, f[i][k]);
             now = max(now, f[i - ][j]);
             f[i][j] = c[i][j] + now;
         }
         //rep(i, 1, n){ rep(j, 1, m) printf("%d ", f[i][j]); putchar(10);}
         printf("%d\n", f[n][m]);
     }
 
     return ;
 
 }

　　还有就是记忆化搜索，很裸的一道题。 HDU1579

　　题目也说了大数据时直接递归会超时，所以除了记忆化搜索，别无选择。

　　

 #include <bits/stdc++.h>
 
 using namespace std;
 
 #define REP(i,n)                for(int i(0); i <  (n); ++i)
 #define rep(i,a,b)              for(int i(a); i <= (b); ++i)
 #define dec(i,a,b)              for(int i(a); i >= (b); --i)
 #define for_edge(i,x)           for(int i = H[x]; i; i = X[i])
 
 #define LL      long long
 #define ULL     unsigned long long
 #define MP      make_pair
 #define PB      push_back
 #define FI      first
 #define SE      second
 #define INF     1 << 30
 
 const int N     =          +       ;
 const int M     =           +       ;
 const int Q     =            +       ;
 const int A     =              +       ;
 
 int f[A][A][A];
 int x, y, z;
 
 int cal(int x, int y, int z){
     if (x <=  || y <=  || z <= ) return ;
     if (x >  || y >  || z > ) return f[][][] = cal(, , );
     if (f[x][y][z] != -) return f[x][y][z];
     if (x < y && y < z) return f[x][y][z] = cal(x, y, z - ) + cal(x, y - , z - ) - cal(x, y - , z);
     else return f[x][y][z] = cal(x - , y, z) + cal(x - , y - , z) + cal(x - , y, z - ) - cal(x - , y - , z - );
 }
 
 int main(){
 #ifndef ONLINE_JUDGE
     freopen("test.txt", "r", stdin);
     freopen("test.out", "w", stdout);
 #endif
 
     memset(f, -, sizeof f);
     while (~scanf("%d%d%d", &x, &y, &z)){
         if (x == - && y == - && z == -) break;
         printf("w(%d, %d, %d) = %d\n", x, y, z, cal(x, y, z));
     }
 
     return ;
 
 }

　　还有一道并查集裸题……HDU1232

　　直接YY一个就A了……

　　

 #include <bits/stdc++.h>
 
 using namespace std;
 
 #define REP(i,n)                for(int i(0); i <  (n); ++i)
 #define rep(i,a,b)              for(int i(a); i <= (b); ++i)
 #define dec(i,a,b)              for(int i(a); i >= (b); --i)
 #define for_edge(i,x)           for(int i = H[x]; i; i = X[i])
 
 #define LL      long long
 #define ULL     unsigned long long
 #define MP      make_pair
 #define PB      push_back
 #define FI      first
 #define SE      second
 #define INF     1 << 30
 
 const int N     =          +       ;
 const int M     =           +       ;
 const int Q     =            +       ;
 const int A     =              +       ;
 
 int f[N];
 int n, m;
 int x, y;
 
 int getfather(int x){ return f[x] ? f[x] = getfather(f[x]) : x;}
 
 int main(){
 #ifndef ONLINE_JUDGE
     freopen("test.txt", "r", stdin);
     freopen("test.out", "w", stdout);
 #endif
 
     while (~scanf("%d", &n), n){
         scanf("%d", &m);
         memset(f, , sizeof f);
         while (m--){
             scanf("%d%d", &x, &y);
             int fa = getfather(x), fb = getfather(y);
             if (fa != fb) f[fa] = fb;
         }
 
         int ans = ;
         rep(i, , n - ) rep(j, i + , n){
             int fa = getfather(i), fb = getfather(j);
             if (fa != fb){
                 f[fa] = fb;
                 ++ans;
             }
             if (ans >= n) break;
         }
 
         printf("%d\n", ans);
     }
 
     return ;
 
 }

　　剩下的一些水题都不好意思拿出来……之后要加大难度了……

　　The End.