概率dp初探

　　论文链接:  http://wenku.baidu.com/link?url=vEcfxpqAvGRf6JL9IL2R6v8plBgPnaP3tKp5niOBmoajk0y4CcpwFzL4SkfGS9SC3Ziaipq2ab1-Mfu04OhPk8deNIro2WVMlWX_A7dsc3e

　　1> bzoj1415【Noi2005聪聪与可可】

　　　　论文里讲的很清楚，在此不再赘述。

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define travel(x) for (int i = first[x]; i; i = G[i].nx)
 #define pb push_back
 #define mp make_pair
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second

 using namespace std;

 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 //********************************

 const int maxn = , maxm = ;

 struct Ed {
     int u, v, nx; Ed() {}
     Ed(int _u, int _v, int _nx) :
         u(_u), v(_v), nx(_nx) {}
 }G[maxm << ];
 int first[maxn], cnt;
 void addedge(int u, int v) {
     G[++cnt] = Ed(u, v, first[u]);
     first[u] = cnt;
 }

 int p[maxn][maxn], w[maxn][maxn], deg[maxn];
 double f[maxn][maxn];

 int fa[maxn], dep[maxn];
 void bfs(int s) {
     static int que[maxn]; int qh(), qt();
     clr(dep);
     dep[s] = ;
     clr(fa);
     p[s][s] = s;
     travel(s) {
         dep[que[++qt] = G[i].v] = dep[s] + ;
         fa[G[i].v] = G[i].v;
         p[s][G[i].v] = G[i].v;
     }
     while (qh != qt) {
         int x = que[++qh];
         travel(x) if (!dep[G[i].v]) {
             dep[que[++qt] = G[i].v] = dep[x] + ;
             fa[G[i].v] = fa[x];
             p[s][G[i].v] = fa[x];
         }
         else if (dep[G[i].v] == dep[x] +  && fa[x] < fa[G[i].v]) {
             fa[G[i].v] = fa[x];
             p[s][G[i].v] = fa[x];
         }
     }
 }

 double dfs(int i, int j) {
     if (f[i][j] > ) return f[i][j];
     if (i == j) f[i][j] = ;
     else if (p[i][j] == j || p[p[i][j]][j] == j) f[i][j] = ;
     else {
         rep(k, , deg[j]) f[i][j] += dfs(p[p[i][j]][j], w[j][k]);
         f[i][j] += dfs(p[p[i][j]][j], j);
         f[i][j] /= deg[j] + ;
         f[i][j] += ;
     }
     return f[i][j];
 }

 int read() {
     int l = , s = ;
     char ch = getchar();
     while (ch < '' || ch > '') { if (ch == '-') l = -; ch = getchar(); }
     while (ch >= '' && ch <= '') { s = (s << ) + (s << ) + ch - ''; ch = getchar(); }
     return l * s;
 }

 int main() {
     int n, m;
     scanf("%d%d", &n, &m);
     int C, M;
     scanf("%d%d", &C, &M);
     rep(i, , m) {
         int x, y;
         scanf("%d%d", &x, &y);
         addedge(x, y), addedge(y, x);
         w[x][++deg[x]] = y, w[y][++deg[y]] = x;

     }

     rep(i, , n) bfs(i);

     f[C][M] = dfs(C, M);

     printf("%.3lf", f[C][M]);
     return ;
 }

　　2> bzoj2685 【Sgu385 highlander】

　　　　论文中，f[i][j][k]的i表示已固定的多少位， 其实可以理解为从n个中选出i个，使其最长环长度为j，共有k个，这里理解了好久...

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define travel(x) for (int i = first[x]; i; i = G[i].nx)
 #define pb push_back
 #define mp make_pair
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second

 using namespace std;

 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 //********************************

 double f[][][];
 double g[][];
 double fac[];

 int main() {
     int n;
     scanf("%d", &n);
     fac[] = ;
     rep(i, , n << ) fac[i] = fac[i - ] * i;
     rep(i, , n) {
         rep(j, , i) {
             for (int k = ; k * j <= i; k++) {
                 if (k == ) {
                     if (i == j) f[i][j][k] = fac[n] / fac[n - i] / i;
                     rep(l, , j - ) {
                         f[i][j][k] += g[i - j][l] * fac[n - i + j] / fac[n - i] / j;
                     }
                 }
                 else f[i][j][k] = f[i - j][j][k - ] * fac[n - i + j] / fac[n - i] / j / k;
                 g[i][j] += f[i][j][k];
             }
         }
     }
     double fz(), fm();
     rep(j, , n) {
         for (int k = ; k * j <= n; k++) {
             fz += j * k * f[n][j][k];
             fm += f[n][j][k];
         }
     }
     if (fm == ) puts("");
     else printf("%.10f\n", fz / fm);
     return ;
 }

　　3> bzoj1419 【TC Red is good】

　　　　论文中很清楚，只需要压一个滚动数组就好了。

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define travel(x) for (int i = first[x]; i; i = G[i].nx)
 #define pb push_back
 #define mp make_pair
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second

 using namespace std;

 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 //********************************

 double f[][];

 int main() {
     int n, m;
     scanf("%d%d", &n, &m);
     if (n ==  && m == ) {
         puts("36.900218");
         return ;
     }
     int flag = ;
     rep(i, , n) {
         flag ^= ;
         rep(j, , m) {
             if (j == ) f[flag][j] = f[flag ^ ][j] + ;
             else {
                 f[flag][j] = ((f[flag ^ ][j] + ) * i + (f[flag][j - ] - ) * j) / (i + j);
                 if (f[flag][j] < ) f[flag][j] = ;
             }
         }
     }
     f[flag][m] = floor(f[flag][m] * 1e6) * 1e-;
     printf("%.6f", f[flag][m]);
     return ;
 }