GYM101147-K. Touristic Trip-概率DP+条件概率

理解了Q巨的代码。模写了一发。

 dp数组含义为在第j个城市发明信片序列第i张的概率。

按照发放顺序转移

对于条件概率的分子，我么可以知道在这个dp数组里面只能从dp[Q][Z]转移过去。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
double dp[30][30];
double path[30][30];
double send[30][30];
int n[30];
int main()
{
    int T;
    freopen("trip.in","r",stdin);
    cin >>T;
    while(T--){
        int N,M,K,Q,Z;
        cin >> N >> M >> K >> Q >> Z;
        Q++;
        Z++;
        for (int i = 1; i<=N; i++){
            for (int j = 1;  j<=N; j++)
                cin >> path[i][j];
        }
        for (int i = 1; i<=N; i++){
            for (int j = 1; j<=M ; j++)
                cin >> send[i][j];
        }
        for (int i = 1; i<=K; i++){cin >> n[i];n[i]++;}
        memset(dp,0,sizeof(dp));
        dp[1][1] = send[1][n[1]];
        for (int m = 2; m<=K; m++)
            for (int now = 1; now<=N; now++ ){
                for (int f = 1; f <=N; f++)
                    dp[m][now] += dp[m-1][f]*path[f][now];
            dp[m][now]*=send[now][n[m]];
            }
        double f1 = 0;
        for (int i = 1; i<=N; i++)
            f1+=dp[K][i];
        memset(dp,0,sizeof(dp));
        if (Q!=1||1==Z)dp[1][1] = send[1][n[1]];
        for (int m = 2; m<=K; m++)
            for (int now = 1; now<=N; now++ ){
                for (int f = 1; f <=N; f++)
                    dp[m][now] += dp[m-1][f]*path[f][now];
            dp[m][now]*=send[now][n[m]];
            if (m==Q&&now!=Z) dp[m][now] = 0;
            }
        double f2 = 0;
        for (int i = 1; i<=N; i++)
            f2+=dp[K][i];
        printf("%.3f
",f2/f1);
    }
    return 0;
}