【题目链接】

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 【算法】

          概率DP

          先跑一遍floyed，求出每个教室之间的最短路径，存在数组dist[][]中，时间复杂度O(V^3)

          设计状态，f[i][j][k]表示当前选到第i个教室，已经选了j个教室，当前这个教室选不选（0..1)

          那么，状态转移方程是什么呢？

          假设当前选到第i个教室，已经选了j个教室，那么，如果不选这个教室，则状态转移方程为

          f[i][j][0] = max{f[i-1][j][0]+dist[c[i-1]][c[i]],f[i-1][j][1]+dist[c[i-1]][c[i]]*(1-k[i-1])+dist[d[i-1]][c[i]]*k[i-1]}

 

      如果选这个教室，则状态转移方程是

      f[i][j][1] = min{f[i-1][j-1][0]+dist[c[i-1]][d[i]]*k[i]+dist[c[i-1]][c[i]]*(1-k[i],f[i-1][j-1][1]+dist[d[i-1]][d[i]]*k[i-1]*k[i]+dist[c[i-1]][d[i]]*(1-k[i-1])*k[i]+dist[c[i-1]][c[i]]*(1-k[i-1])*(1-k[i])+dist[d[i-1]][c[i]]*k[i-1]*(1-k[i])}

 

      于是这道题便迎刃而解了！

      【代码】

   

#include<bits/stdc++.h>
using namespace std;
#define MAXN 2000
#define MAXV 300

int i,j,k,n,m,v,e,a,b,w;
int dist[MAXV+10][MAXV+10],c[MAXN+10],d[MAXN+10];
double P[MAXN+10],dp[MAXN+10][MAXN+10][2];
double ans = 2e9;

template <typename T> inline void read(T &x) {
        int f=1; x=0;
        char c = getchar(); 
        for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
        for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
        x *= f;
}

template <typename T> inline void write(T x) {
    if (x < 0) { putchar('-'); x = -x; }
    if (x > 9) write(x/10);
    putchar(x%10+'0');    
}

template <typename T> inline void writeln(T x) {
    write(x);
    puts("");    
}

int main() {
        
        read(n); read(m); read(v); read(e);
        for (i = 1; i <= v; i++) {
                for (j = 1; j <= v; j++) {
                        if (i != j)
                                dist[i][j] = 2e9;
                }
        }
         for (i = 1; i <= n; i++) read(c[i]);
        for (i = 1; i <= n; i++) read(d[i]);
        for (i = 1; i <= n; i++) cin >> P[i];
        for (i = 1; i <= e; i++) {
                read(a); read(b); read(w);
                dist[a][b] = min(dist[a][b],w);
                dist[b][a] = min(dist[b][a],w);    
        }
    
        for (k = 1; k <= v; k++) {
                for (i = 1; i <= v; i++) {
                        if (i == k) continue;
                        if (dist[i][k] == 2e9) continue;
                        for (j = 1; j <= v; j++) {
                                if (dist[k][j] == 2e9) continue;
                                if ((i == j) || (k == j)) continue;
                                dist[i][j] = min(dist[i][j],dist[i][k]+dist[k][j]);
                        }
                }
        }    
        
        for (i = 1; i <= n; i++) {
                for (j = 0; j <= m; j++) {
                        dp[i][j][0] = dp[i][j][1] = 2e9;
                }    
        }
        
        dp[1][0][0] = 0; dp[1][1][1] = 0;
        for (i = 2; i <= n; i++) {
                dp[i][0][0] = dp[i-1][0][0] + dist[c[i-1]][c[i]];
                for (j = 1; j <= min(i,m); j++) {
                        dp[i][j][0] = min(dp[i-1][j][0]+dist[c[i-1]][c[i]],dp[i-1][j][1]+dist[c[i-1]][c[i]]*(1.0-P[i-1])+dist[d[i-1]][c[i]]*P[i-1]);
                        dp[i][j][1] = min(dp[i-1][j-1][0]+dist[c[i-1]][d[i]]*P[i]*1.0+dist[c[i-1]][c[i]]*(1.0-P[i]),
                                                            dp[i-1][j-1][1]+
                                                            dist[d[i-1]][d[i]]*P[i-1]*P[i]*1.0+ 
                                                            dist[c[i-1]][d[i]]*(1.0-P[i-1])*P[i]+ 
                                                            dist[c[i-1]][c[i]]*(1.0-P[i-1])*(1.0-P[i])+ 
                                                                dist[d[i-1]][c[i]]*P[i-1]*(1-P[i])*1.0); 
                }    
        }
        
        for (i = 0; i <= m; i++) ans = min(ans,min(dp[n][i][0],dp[n][i][1]));
        cout<< fixed << setprecision(2) << ans << endl;
        
        return 0;
    
}