https://cn.vjudge.net/problem/ZOJ-3605

Alice and Bob are playing a game. This game is played with several identical pots and one marble. When the game starts, Alice puts the pots in one line and puts the marble in one of the pots. After that, Bob cannot see the inside of the pots. Then Alice makes a sequence of swappings and Bob guesses which pot the marble is in. In each of the swapping, Alice chooses two different pots and swaps their positions.

Unfortunately, Alice's actions are very fast, so Bob can only catch k of m swappings and regard these k swappings as all actions Alice has performed. Now given the initial pot the marble is in, and the sequence of swappings, you are asked to calculate which pot Bob most possibly guesses. You can assume that Bob missed any of the swappings with equal possibility.
Input

There are several test cases in the input file. The first line of the input file contains an integer N (N ≈ 100), then N cases follow.

The first line of each test case contains 4 integers n, m, k and s(0 < s ≤ n ≤ 50, 0 ≤ k ≤ m ≤ 50), which are the number of pots, the number of swappings Alice makes, the number of swappings Bob catches and index of the initial pot the marble is in. Pots are indexed from 1 to n. Then m lines follow, each of which contains two integersai and bi (1 ≤ ai, bi ≤ n), telling the two pots Alice swaps in the i-th swapping.
Outout

For each test case, output the pot that Bob most possibly guesses. If there is a tie, output the smallest one.
Sample Input

3
3 1 1 1
1 2
3 1 0 1
1 2
3 3 2 2
2 3
3 2
1 2

Sample Output

2
1
3
一共有 n 个杯子，有一个石头开始是放在第 s 号杯子里，然后交换这些杯子 m 次，但是只能记住其中的 k 次，每次交换没有看到交换的概率相同，求最终猜的那个人最可能猜的是哪号杯子里有石头。

dp[i][j][k]代表交换了i次，看到了j次，石子在k位置

https://blog.csdn.net/u012860063/article/details/45199239

#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
ll dp[55][55][55];
int a[55],b[55]; 
int n,m,k,s;
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
    	scanf("%d%d%d%d",&n,&m,&k,&s);
    	memset(dp,0,sizeof(dp));
    	dp[0][0][s]=1;
    	for(int i=1;i<=m;i++){
    	    scanf("%d%d",&a[i],&b[i]);
	}
	for(int i=1;i<=m;i++){
	    dp[i][0][s]=1;
	    for(int j=1;j<=min(i,k);j++){
		dp[i][j][a[i]]=dp[i-1][j-1][b[i]];//看到石子交换
		dp[i][j][b[i]]=dp[i-1][j-1][a[i]];
		for(int p=1;p<=n;p++) {	//没看到石子交换
		    dp[i][j][p]+=dp[i-1][j][p];//没看到交换
		    if(p!=a[i]&&p!=b[i]) //看到交换，但没石子 
		    dp[i][j][p]+=dp[i-1][j-1][p];
		}
	     }
	}
	int ans=1;
        for(int i=2;i<=n;i++){
            if(dp[m][k][i]>dp[m][k][ans]) ans=i;
	} 
	printf("%d\n",ans);
    }
    return 0;
}