Ponds HDU - 5438 （拓扑排序）

最新推荐文章于 2022-09-06 15:27:48 发布

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本篇介绍了一个关于移除池塘的算法问题，Betty需要移除一些池塘来节省开支，并遵循特定规则。文章通过使用拓扑排序的方法解决此问题，实现了有效移除并计算剩余奇数个池塘组件的价值总和。

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 vv.

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)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)p(1≤p≤104) which represents the number of ponds she owns, and the other is the number m(1≤m≤105)m(1≤m≤105) which represents the number of pipes.

The next line contains pp numbers v1,…,vpv1,…,vp, where vi(1≤vi≤108)vi(1≤vi≤108) indicating the value of pond ii.

Each of the last mm lines contain two numbers aa and bb, which indicates that pond aa and pond bb 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

后来想到了拓扑排序，一个一个删掉吧，简单题

#include<bits/stdc++.h>
#define INF 1000000000
#define N 10004
using namespace std;
int n,m;
bool vis[N];
int a[N];
int d[N];
vector<int> g[N];
long long ans;
long long temp;
int num;
bool mark[N];
void dfs(int u)
{
    num++;
    temp+=a[u];
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i];
        if(mark[v])
            continue;
        if(!vis[v])
        {
            vis[v]=true;
            dfs(v);
        }
    }
}
queue<int> que;
void process()
{
    for(int i=1;i<=n;i++)
    {
        if(d[i]<2)
        {
           que.push(i);
            mark[i]=true;
        }
    }
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        for(int i=0;i<g[u].size();i++)
        {
            int v=g[u][i];
            d[v]--;
            if(d[v]==1&&!mark[v])
            {
                //cout<<v<<endl;
                que.push(v);
                mark[v]=true;
            }
        }
    }
}
void init()
{
    for(int i=1;i<=n;i++)
        g[i].clear();
    memset(d,0,sizeof(d));
    memset(vis,false,sizeof(vis));
    memset(mark,false,sizeof(mark));
}
int main()
{
    int x,y,w;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        while(m--)
        {
            scanf("%d%d",&x,&y);
            g[x].push_back(y);
            g[y].push_back(x);
            d[x]++;
            d[y]++;
        }

        process();

        for(int i=1;i<=n;i++)
        {
            if(!vis[i]&&!mark[i])
            {
                vis[i]=true;
                temp=0;
                num=0;
                dfs(i);
                if(num&1)
                    ans+=temp;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}