HDU 5631 Rikka with Graph / UVALive - 7240 Minimum Cut-Cut（删边操作，BestCoder）

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本文介绍了一道关于图论中最小割问题的竞赛题目，通过对给定图进行边的删除操作来保持图的连通性，并计算所有可行的删除组合数量。文章详细阐述了解题思路及实现代码。

Rikka with Graph（进入答题）

UVALive - 7240 Minimum Cut-Cut（答题）

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1031 Accepted Submission(s): 491

Problem Description

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?

Input

The first line contains a number

T(T≤30) ——The number of the testcases.

For each testcase, the first line contains a number

n(n≤100) .

Then n+1 lines follow. Each line contains two numbers

u,v , which means there is an edge between u and v.

Output

For each testcase, print a single number.

Sample Input

Sample Output

Source

BestCoder Round #73 (div.2)

题意：

就是给定一个图，删除一条边或者两条边之后，仍然能够保持连通的操作有多少种。

思路：

第一次接触删边操作，自然参考了题解，但愿自己的注释没写错。

AC代码：

#include<stdio.h>
#define MYDD 1103

int n,pre[MYDD];//记录父节点
struct node { //记录边信息的结构体
	int u;//起点
	int v;//终点
	int flag;//标记是否已经使用
} dd[MYDD];

void init(int x) {
	for(int j=1; j<=x; j++)
		pre[j]=j;
}

int find(int x) {
	return x==pre[x]? x:find(pre[x]);
}

void combine(int x,int y) {
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
		pre[fx]=fy;
}

bool Delete() { //删除边的操作
	init(n);
	for(int j=0; j<=n; j++) {
		if(dd[j].flag)
			combine(dd[j].u,dd[j].v);
	}

	int sum=0;//统计树的个数，即是否连通
	for(int j=1; j<=n; j++) {
		if(pre[j]==j)//根节点等于本身就是一棵树
			sum++;
		if(sum>1) {//存在大于 1 个的树 
			return false;//不成立 
		}
	}
	return true;//成立 
}

int main() {
	int t;
	scanf("%d",&t);
	while(t--) {
		scanf("%d",&n);
		for(int j=0; j<=n; j++) {
			scanf("%d%d",&dd[j].u,&dd[j].v);
			dd[j].flag=1;
		}

		int ans=0;//记录删除边之后满足的答案 
		for(int j=0; j<=n; j++) {
			dd[j].flag=0;
			if(Delete())//删除一条边
				ans++;
			for(int k=j+1; k<=n; k++) {
				dd[k].flag=0;
				if(Delete())//删除两条边的操作
					ans++;
				dd[k].flag=1;//边的还原
			}
			dd[j].flag=1;//边的还原，为了下一次删边保证原图完整 
		}

		printf("%d\n",ans);
	}
	return 0;
}

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