标签：
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 ?
2 5 1 2 2 3 3 4 4 5
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;
}
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原文地址：http://www.cnblogs.com/Blackops/p/5815104.html
