Find the Marble

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 ofm 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 integerN (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 integersa_i and b_i (1 ≤ a_i, b_i ≤ 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

题目大意：N个容器，M次两两交换。当中K次是能够知道的，一開始珠子放在当中一个容器里。问你交换完以后，珠子在哪个容器的概率最大

思路：用DP[m][k][n] 表示 m次交换，知道了当中的k次，结尾为n的方案数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using  namespace std;
typedef long long ll;
const  int maxn = 60;
int n,m,k,s;
int A[maxn],B[maxn];
ll dp[maxn][maxn][maxn];
int main(){

    int ncase;
    cin >> ncase;
    while(ncase--){
        scanf("%d%d%d%d",&n,&m,&k,&s);
        for(int i = 1; i <= m; i++){
            scanf("%d%d",&A[i],&B[i]);
        }
        memset(dp,0,sizeof dp);
        dp[0][0][s] = 1;
        for(int i = 1; i <= m; i++){
            dp[i][0][s] = 1;
            for(int j = 1; j <= i&&j <= k; j++){
                dp[i][j][A[i]] = dp[i-1][j-1][B[i]];
                dp[i][j][B[i]] = dp[i-1][j-1][A[i]];
                for(int d = 1; d <= n; d++){
                    dp[i][j][d] += dp[i-1][j][d];
                    if(d!=A[i]&&d!=B[i]){
                        dp[i][j][d] += dp[i-1][j-1][d];
                    }
                }
            }
        }
        int idx = 1;
        for(int i = 2; i <= n; i++){
            if(dp[m][k][i] > dp[m][k][idx]){
                idx = i;
            }
        }
        cout<<idx<<endl;
    }
    return 0;
}