2015ACM/ICPC Asia Regional Changchun Online /HDU 5438 图

                                               Ponds

                                  Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                       

Problem Description

Betty owns a lot of ponds, some of them are connected with other ponds by pipes, and there will not be more than one pipe between two ponds. Each pond has a value v.

Now Betty wants to remove some ponds because she does not have enough money. But each time when she removes a pond, she can only remove the ponds which are connected with less than two ponds, or the pond will explode.

Note that Betty should keep removing ponds until no more ponds can be removed. After that, please help her calculate the sum of the value for each connected component consisting of a odd number of ponds

 

Input

The first line of input will contain a number T(1≤T≤30) which is the number of test cases.

For each test case, the first line contains two number separated by a blank. One is the number p(1≤p≤104) which represents the number of ponds she owns, and the other is the number m(1≤m≤105) which represents the number of pipes.

The next line contains p numbers v1,...,vp, where vi(1≤vi≤108) indicating the value of pond i.

Each of the last m lines contain two numbers a and b, which indicates that pond a and pond b are connected by a pipe.

 

Output

For each test case, output the sum of the value of all connected components consisting of odd number of ponds after removing all the ponds connected with less than two pipes.

 

Sample Input

1 7 7 1 2 3 4 5 6 7 1 4 1 5 4 5 2 3 2 6 3 6 2 7

 

Sample Output

21

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

 

姿势题

题意：给你n点,m边，每次删除其连通边数小于2的点，直到不能删除，问你最后所剩点的价值为多少

//1085422276
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define memfy(a) memset(a,-1,sizeof(a))
#define TS printf("111111
");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define mod 1000000007
#define maxn 1000006
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//****************************************

int n,m;
ll sum;
int s,vis[maxn*2],v[maxn],out[maxn];
vector<int >G[maxn];
void init()
{
    FOR(i,1,n)G[i].clear();
    mem(vis);
    mem(out);
}
void dfs(int x)
{
    vis[x]=1;
    s++;
    sum+=v[x];
    for(int i=0;i<G[x].size();i++)
    {
        if(!vis[G[x][i]])
        {
            dfs(G[x][i]);
        }
    }
}
int main()
{
    int T=read();
    while(T--)
    {
        init();
        scanf("%d%d",&n,&m);
        FOR(i,1,n)
        {
            scanf("%d",&v[i]);
        }
        int a,b;
        FOR(i,1,m)
        {
            scanf("%d%d",&a,&b);
            G[a].push_back(b);
            G[b].push_back(a);
            out[a]++;
            out[b]++;
        }
        ll ans=0;
        queue<int >q;
        FOR(i,1,n)
        {
            if(out[i]<=1)q.push(i);
            ans+=v[i];
        }
        while(!q.empty())
        {
            int k=q.front();
            q.pop();
            vis[k]=1;
            ans-=v[k];
            for(int i=0;i<G[k].size();i++)
            {
                out[G[k][i]]--;
               if(out[G[k][i]]<=1&&!vis[G[k][i]])q.push(G[k][i]);
            }
        }
       FOR(i,1,n)
       {
           sum=0; s=0;
           if(!vis[i])dfs(i);
           if(s%2==0)ans-=sum;
       }
       cout<<ans<<endl;
    }
    return 0;
}