题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5438

Ponds

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

Total Submission(s): 2837    Accepted Submission(s): 891

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 valuev.

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 numberp(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

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题目大意：给出一个无向图，每一个节点都有一定的权值，最后要求输出每一个奇数圈内的权值和。

解题思路：首先 你要考虑怎么，保证它是一个圈。这样就想到采用拓扑排序的方法，去掉所有入度<2的点。再就是判断这个圈是否含有奇数个点，采用深搜的方法将每个点做一次起始点，对其进行搜索计数！

详见代码。

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>

using namespace std;

#define N 10010
#define ll long long

ll val[N],V;
int vis[N];//判断这个点有没有删掉
int Vis[N];//判断这个点有没有做为起点进行搜索过
int indir[N],k;
vector<int>G[N];

void dfs(int st)
{
    k++;
    V+=val[st];
    for (int i=0; i<G[st].size(); i++)
    {
        if (!Vis[G[st][i]]&&!vis[G[st][i]])
        {
            Vis[G[st][i]]=1;
            dfs(G[st][i]);
        }
    }
}

int main()
{
    int t,a,b;
    scanf("%d",&t);
    while (t--)
    {
        int n,m;
        memset(vis,0,sizeof(vis));
        memset(Vis,0,sizeof(Vis));
        memset(val,0,sizeof(val));
        memset(indir,0,sizeof(indir));
        scanf("%d%d",&n,&m);
        for (int i=1; i<=n; i++)
            G[i].clear();
        for (int i=1; i<=n; i++)
            scanf("%lld",&val[i]);
        for (int i=1; i<=m; i++)
        {
            scanf("%d%d",&a,&b);
            G[a].push_back(b);
            G[b].push_back(a);
            indir[a]++;
            indir[b]++;
        }
        while (1)
        {
            int i;
            for (i=1; i<=n; i++)
            {
                if (indir[i]<2&&!vis[i])
                {
                    vis[i]=1;
                    for (int j=0; j<G[i].size(); j++)
                    {
                        if (!vis[G[i][j]])
                            indir[G[i][j]]--;
                    }
                    break;
                }
            }
            if (i>n)
                break;
        }
        ll ans=0;
        for (int i=1; i<=n; i++)
        {
            if (!Vis[i]&&!vis[i])
            {
                k=0;
                V=0;
                Vis[i]=1;
                dfs(i);
                if (k%2==1)
                    ans+=V;
            }
        }
        printf ("%lld\n",ans);
    }
    return 0;
}