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 = 310, M = 2010;
const double INF = 1.0 * 0x3f3f3f3f;

inline void read(int &x) {
    x = 0;
    char c = getchar();
    while(c > '9' || c < '0') {
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + c - 48;
        c = getchar();
    }
    return;
}

int G[N][N], a[M], b[M];
double k[M], f[M][M][2];

int main() {
    int n, m, v, e;
    read(n);
    read(m);
    read(v);
    read(e);
    for(int i = 1; i <= n; i++) {
        read(a[i]);
    }
    for(int i = 1; i <= n; i++) {
        read(b[i]);
    }
    for(int i = 1; i <= n; i++) {
        scanf("%lf", &k[i]);
    }
    int x, y, z;
    memset(G, 0x3f, sizeof(G));
    for(int i = 1; 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 = 1; i <= v; i++) {
        G[i][i] = 0;
    }
    for(int p = 1; p <= v; p++) {
        for(int i = 1; i <= v; i++) {
            for(int j = 1; j <= v; j++) {
                G[i][j] = std::min(G[i][j], G[i][p] + G[p][j]);
            }
        }
    }

    for(int i = 1; i <= n; i++) {
        for(int j = 0; j <= m; j++) {
            f[i][j][0] = f[i][j][1] = INF;
        }
    }
    f[1][0][0] = f[1][1][1] = 0;
    for(int i = 2; i <= n; i++) {
        for(int j = 0; j <= m; j++) {
            f[i][j][0] = std::min(f[i - 1][j][0] + G[a[i - 1]][a[i]],
                                  f[i - 1][j][1]
                                  + G[b[i - 1]][a[i]] * k[i - 1]
                                  + G[a[i - 1]][a[i]] * (1 - k[i - 1]));
            if(j) {
                f[i][j][1] = std::min(f[i - 1][j - 1][0]
                                      + G[a[i - 1]][b[i]] * k[i]
                                      + G[a[i - 1]][a[i]] * (1 - k[i]),
                                      f[i - 1][j - 1][1]
                                      + G[b[i - 1]][b[i]] * k[i - 1] * k[i]
                                      + G[b[i - 1]][a[i]] * k[i - 1] * (1 - k[i])
                                      + G[a[i - 1]][b[i]] * (1 - k[i - 1]) * k[i]
                                      + G[a[i - 1]][a[i]] * (1 - k[i - 1]) * (1 - k[i]));
            }
        }
    }

    double ans = INF;

    for(int i = 0; i <= m; i++) {
        ans = std::min(ans, f[n][i][0]);
        ans = std::min(ans, f[n][i][1]);
    }
    printf("%.2lf", ans);

    return 0;
}

我一开始把ans初值设的是0...