洛谷P1850 换教室

令人印象深刻的状态转移方程...

f[i][j][0/1]表示前i个换j次，第i次是否申请时的期望。

注意可能有重边，自环。

转移要分类讨论，距离是上/这次成功/失败的概率乘相应的路程。

从上次的0/1中取min

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 const int N = , M = ;
 const double INF = 1.0 * 0x3f3f3f3f;

 inline void read(int &x) {
     x = ;
     char c = getchar();
     while(c > '' || c < '') {
         c = getchar();
     }
     while(c >= '' && c <= '') {
         x = (x << ) + (x << ) + c - ;
         c = getchar();
     }
     return;
 }

 int G[N][N], a[M], b[M];
 double k[M], f[M][M][];

 int main() {
     int n, m, v, e;
     read(n);
     read(m);
     read(v);
     read(e);
     for(int i = ; i <= n; i++) {
         read(a[i]);
     }
     for(int i = ; i <= n; i++) {
         read(b[i]);
     }
     for(int i = ; i <= n; i++) {
         scanf("%lf", &k[i]);
     }
     int x, y, z;
     memset(G, 0x3f, sizeof(G));
     for(int i = ; i <= e; i++) {
         read(x);
         read(y);
         read(z);
         if(G[x][y]) {
             G[y][x] = G[x][y] = std::min(G[x][y], z);
         }
         else {
             G[x][y] = G[y][x] = z;
         }
     }

     /// floyd
     for(int i = ; i <= v; i++) {
         G[i][i] = ;
     }
     for(int p = ; p <= v; p++) {
         for(int i = ; i <= v; i++) {
             for(int j = ; j <= v; j++) {
                 G[i][j] = std::min(G[i][j], G[i][p] + G[p][j]);
             }
         }
     }

     for(int i = ; i <= n; i++) {
         for(int j = ; j <= m; j++) {
             f[i][j][] = f[i][j][] = INF;
         }
     }
     f[][][] = f[][][] = ;
     for(int i = ; i <= n; i++) {
         for(int j = ; j <= m; j++) {
             f[i][j][] = std::min(f[i - ][j][] + G[a[i - ]][a[i]],
                                   f[i - ][j][]
                                   + G[b[i - ]][a[i]] * k[i - ]
                                   + G[a[i - ]][a[i]] * ( - k[i - ]));
             if(j) {
                 f[i][j][] = std::min(f[i - ][j - ][]
                                       + G[a[i - ]][b[i]] * k[i]
                                       + G[a[i - ]][a[i]] * ( - k[i]),
                                       f[i - ][j - ][]
                                       + G[b[i - ]][b[i]] * k[i - ] * k[i]
                                       + G[b[i - ]][a[i]] * k[i - ] * ( - k[i])
                                       + G[a[i - ]][b[i]] * ( - k[i - ]) * k[i]
                                       + G[a[i - ]][a[i]] * ( - k[i - ]) * ( - k[i]));
             }
         }
     }

     double ans = INF;

     for(int i = ; i <= m; i++) {
         ans = std::min(ans, f[n][i][]);
         ans = std::min(ans, f[n][i][]);
     }
     printf("%.2lf", ans);

     return ;
 }

AC代码

我一开始把ans初值设的是0...