UPC 1053 Mysterious Treasure （记忆化搜索）

这个题开始没想到全边的测试数据，不用记忆化搜索会 50个点800条边会搜索2*10^7次，然后改成记忆化搜索，暴力删边，昨天有点乱，结束了立马想到暴力删边哪里没解决，生气

Description

WNJXYK and DIDIDI is playing a game. DIDIDI draws a directed graph G on the paper which contains n points, m directed edges and no cycle. WNJXYK starts from point 1. For every round, WNJXYK will randomly select one of the directed edges which start from current point with equal possibility, and go to next point by this edge. And the game will continue until such edge doesn’t exist. DIDIDI will place the treasure on point n, if WNJXYK go through this point, he can get the treasure. WNJXYK have one chance to delete one edge(he can also don’t choose to delete), that he can increase the probability of getting a treasure. Your assignment is to calculate the probability of WNJXYK getting the treasure in the optimal condition.

 

Input

The first line of input contains a positive integer T telling you there are T test cases followed.

For each test case, the first line contains two integers n , m, indicating the number of points ,the number of edges.

Then the following are m lines, each of which contains two integers x and y, indicating there is a edge from x to y.

It is guaranteed that there does not exist multiple edge.

 

Output

For each test case, print a line “Case #x: y”, where x is the case number (starting from 1) and y is the probability of he getting the treasure. (round to six decimal places).

 

Sample Input

24 41 21 31 42 34 51 21 31 42 32 4
Null

 

Sample Output

Case #1: 0.500000
Case #2: 0.750000
#include<bits/stdc++.h>using namespace std;typedef long long  ll;vector<int>g[53];vector<bool>g2[53],vis[53];bool vis1[53];double ans,dp[53],maxn;int n,flag=0;double dfs(int u,double k){    if(vis1[u])return dp[u]*k;    vis1[u]=1;    double tmp=0;    if(u==n)    {        dp[u]=1;        return k;    }    if(g[u].size()==0)    {        dp[u]=0;        return 0;    }    double t1=0;    for(int i=0; i<g[u].size(); i++)    {        if(vis[u][i]==0)continue;        t1=dfs(g[u][i],k/(u==flag?g[u].size()-1:g[u].size()));        if(t1>0)g2[u][i]=1;        tmp+=t1;    }    dp[u]=tmp/k;    return tmp;}int main(){    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    int t,cnt=1,m,x,y;    scanf("%d",&t);    while(t--)    {        maxn=0;        scanf("%d%d",&n,&m);        memset(vis1,0,sizeof(vis1));        for(int i=1; i<=n; i++)g[i].clear(),g2[i].clear(),dp[i]=0,vis[i].clear();        for(int i=1; i<=m; i++)        {            scanf("%d%d",&x,&y);            if(x==n)continue;            g[x].push_back(y);           vis[x].push_back(1);            g2[x].push_back(0);        }        maxn=max(ans,dfs(1,1));       // printf("%.6f\n",maxn);         for(int i=1; i<n; i++)    {        for(int j=0; j<g[i].size(); j++)        {            if(g2[i][j]==0)            {                memset(vis1,0,sizeof(vis1));                vis[i][j]=0;                flag=i;                memset(dp,0,sizeof(dp));               // cout<<i<<" "<<j<<" "<<dfs(1,1)<<endl;                maxn=max(maxn,dfs(1,1));                vis[i][j]=1;                flag=0;            }        }    }        printf("Case #%d: %.6f\n",cnt++,maxn);    }    return 0;}