B - Rikka with Graph HDU - 5631 (並查集+思維)


As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: 

Yuta has a non-direct graph with n vertices and n+1 edges. Rikka can choose some of the edges (at least one) and delete them from the graph. 

Yuta wants to know the number of the ways to choose the edges in order to make the remaining graph connected. 

It is too difficult for Rikka. Can you help her?

InputThe first line contains a number T(T30)T(T≤30)——The number of the testcases. 

For each testcase, the first line contains a number n(n100)n(n≤100). 

Then n+1 lines follow. Each line contains two numbers u,vu,v , which means there is an edge between u and v.
OutputFor each testcase, print a single number.Sample Input

1
3
1 2
2 3
3 1
1 3

Sample Output

9

 

題意:給出n個點,和n+1條邊,問可以有多少種去掉邊的方法,使去掉邊后整個圖仍然是連通的

 

題解:使用並查集來判斷是否連通,再通過逐個枚舉去掉一條邊和去掉兩條邊的情況,判斷整個圖是否連通,如果是則ans++ 否則ans不變

 

AC代碼:

#include<iostream>
#include<cstdio>

using namespace std;

int s[105], e[105];
int t, n;
int a, b;
int pre[105];

int Find(int r) {
    return pre[r] = pre[r] == r ? r : Find(pre[r]);
}

int check(int a, int b) {
    for (int i = 1; i <= n; i++) {
        pre[i] = i;
    }
    for (int i = 0; i <= n; i++) {
        //與a , b 相連的邊直接去掉,查看是否還能夠全部聯通
        if (i == a || i == b)
            continue;
        int f1 = Find(s[i]), f2 = Find(e[i]);
        if (f1 != f2)
            pre[f1] = f2;
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        if (pre[i] == i)
            cnt++;
        if (cnt > 1)
            return 0;
    }
    return 1;
}
int main() {
    cin >> t;
    while (t--) {
        cin >> n;
        for (int i = 0; i <= n; i++) {
            cin >> s[i] >> e[i];
        }
        int ans = 0;
        //逐個查找,i = j 代表是取一條邊,不等代表是取兩條邊
        //要想全部聯通至少需要n-1條邊
        for (int i = 0; i <= n; i++) {
            for (int j = i; j <= n; j++) {
                ans += check(i, j);
            }
        }
        cout << ans << endl;
    }
    return 0;
}

  

 


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