# UPC 1053 Mysterious Treasure （记忆化搜索）

## 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

2
4 4
1 2
1 3
1 4
2 3
4 5
1 2
1 3
1 4
2 3
2 4


## Sample Output

Case #1: 0.500000
Case #2: 0.750000


## AC 代码

#include<bits/stdc++.h>
using namespace std;
#define eps (1e-7)

const int maxn = 200;
struct node
{
int to;
int next;
} edge[maxn*maxn];
int out[maxn];
double viss[maxn],ans;
bool vis[maxn][maxn];
int delx,dely;

void init()
{
memset(viss,0,sizeof(viss));
memset(out,0,sizeof(out));
memset(vis,false,sizeof(vis));
tot=0;
}

{
edge[tot].to=v;
}

double dfs(int x)
{
double ans=0.0;
if(x==1)        // 起点
ans = 1.0;
else
{
if(fabs(viss[x])>eps)return viss[x];    // 之前计算过的部分
{
int to=edge[i].to;
if(to==delx&&x==dely)continue;
ans+=dfs(to)/out[to];       // 等概率分配
}
}
viss[x]=ans;
return ans;
}

int main()
{
int T;
scanf("%d",&T);
for(int ti=1; ti<=T; ti++)
{
init();
scanf("%d%d",&n,&m);
for(int i=0; i<m; i++)
{
int u,v;
scanf("%d%d",&u,&v);
vis[u][v]=true;
out[u]++;
}
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
{
if(vis[i][j])   // 枚举删除每一条边
{
delx=i,dely=j;
memset(viss,0.0,sizeof(viss));
out[i]--;
ans=max(ans,dfs(n));
out[i]++;
}
}
memset(viss,0.0,sizeof(viss));
delx=dely=-1;
ans=max(ans,dfs(n));    // 不删边的情况
printf("Case #%d: %.6lf\n",ti,ans);
}
return 0;
}