NBUT 1635 Explosion（最小顶点覆盖）

Posted 2020-08-04 Blackops

• [1635] Explosion

• 时间限制: 10000 ms　内存限制: 65535 K
• 问题描述

• there is a country which contains n cities connected by n - 1 roads(just like a tree). If you place TNT in one city, all the roads connect these city will be destroyed, now i want to destroy all the roads with the least number of TNT, can you help me ?

• 输入
• Input starts with an integer T(T <= 500), denoting the number of test case.
  For each test case, first line contains n(1 <= n <= 1000), denoting the number of cities, next n - 1lines following and each line contains two different cities denoting these two cities connect directly. You can assume the input guarantee the relation among cities is a tree.
  
• 输出
• For each test case, print the least number of TNT that i need to destroy all the n - 1 roads.
  
• 样例输入

• 2
  5
  1 2
  2 3
  3 4
  4 5

• 样例输出

• 2

 

题目链接：NBUT 1635

又是一道没人写的水题……由于题目中说like a tree，因此可以归为二分图，然后就套公式，在二分图中最小顶点覆盖数=最大匹配数。（另外这题应该是可以用树形DP做然而并不会……以后再说- -|||）

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1010;
struct edge
{
	int to;
	int pre;
};
edge E[N<<1];
int head[N],ne;
int vis[N],match[N];
void add(int s,int t)
{
	E[ne].to=t;
	E[ne].pre=head[s];
	head[s]=ne++;
}
void init()
{
	CLR(head,-1);
	ne=0;
	CLR(match,-1);
}
int dfs(int now)
{
	for (int i=head[now]; ~i; i=E[i].pre)
	{
		int v=E[i].to;
		if(!vis[v])
		{
			vis[v]=1;
			if(match[v]==-1||dfs(match[v]))
			{
				match[v]=now;
				return 1;
			}
		}
	}
	return 0;
}
int hun(int n)
{
	int r=0;
	for (int i=1; i<=n; ++i)
	{
		CLR(vis,0);
		if(dfs(i))
			++r;
	}
	return r;
}
int main(void)
{
	int tcase,n,a,b,i,j,ans;
	scanf("%d",&tcase);
	while (tcase--)
	{
		init();
		scanf("%d",&n);
		for (i=0; i<n-1; ++i)
		{
			scanf("%d%d",&a,&b);
			add(a,b);
			add(b,a);
		}
		printf("%d\n",hun(n)>>1);
	}
	return 0;
}