期望dp

换教室

• f[i][j][0/1]表示前i个时间点，共申请了j次，第i个时间点否/是进行了申请。

• dis[a][b]表示a教室->b教室的距离

• c[i]表示默认的教室

• d[i]表示更换后的教室

• k[i]表示第i个教室申请成功的概率

【解题思路】

• 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int inf = 0x3fffffff;
double k[2005],f[2005][2005][2];
int n,m,v,e,dis[305][305],d[2005],c[2005];
void work(){
    for(int i = 1; i <= n; i++)
        for(int j = 0; j <= m; j++)
            f[i][j][0] = f[i][j][1] = inf;
    f[0][0][0] = 0;
　　 //可以写成:
    //f[1][0][0] = 0; f[1][1][1] = 0;
    for(int i = 1; i <= n; i++)
        for(int j = 0; j <= m; j++){
            f[i][j][0] = min(f[i-1][j][0] + dis[c[i-1]][c[i]], f[i-1][j][1] +
                        k[i-1]*(dis[d[i-1]][c[i]]) + (1-k[i-1])*dis[c[i-1]][c[i]]);//error:d[i-1]->d[i]
            if(j != 0)
                f[i][j][1] = min(f[i-1][j-1][0] + (1-k[i])*dis[c[i-1]][c[i]] +
                                 k[i]*dis[c[i-1]][d[i]],f[i-1][j-1][1] + k[i]*k[i-1]*dis[d[i-1]][d[i]] + (1-k[i-1])*(1-k[i])*dis[c[i-1]][c[i]] +
                                 k[i-1]*(1-k[i])*dis[d[i-1]][c[i]] + (1-k[i-1])*k[i]*dis[c[i-1]][d[i]]);
        }
}
int main(){
    scanf("%d%d%d%d",&n,&m,&v,&e);
    for(int i = 1; i <= n; i++)
        scanf("%d",&c[i]);
    for(int i = 1; i <= n; i++)
        scanf("%d",&d[i]);
    for(int i = 1; i <= n; i++)
        scanf("%lf",&k[i]);
    for (int i=1;i<=v;i++)
        for (int j=1;j<i;j++)
            dis[j][i]=dis[i][j]=inf;
    for (int i=1;i<=e;i++){
        int a,b,w;
        scanf("%d%d%d",&a,&b,&w);
        dis[a][b]=min(dis[a][b],w);//
        dis[b][a]=dis[a][b];
    }
    for(int k = 1; k <= v; k++)
        for(int i = 1; i <= v; i++)
            for(int j = 1; j < i; j++)
                if(dis[i][k] + dis[k][j] < dis[i][j])
                    dis[i][j] = dis[j][i] = dis[i][k] + dis[k][j] ;
    work();
    double ans = inf;
    for(int i = 0; i <= m; i++)//error:i=0 -> i = 1
        for(int j = 0; j <= 1; j++)
            ans = min(ans, f[n][i][j]);
    printf("%0.2lf",ans);
    return 0;
}

 绿豆蛙的归宿

待定