Ponds

Time Limit: 1 Sec ?

Memory Limit: 256 MB

題目連接

http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1001&cid=621

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

HINT

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題意

給你一個圖，然后要求把度數(shù)小于2的點(diǎn)全部刪去，然后問你奇數(shù)集合的點(diǎn)權(quán)和是多少

注意，你刪去了一個點(diǎn)之后，可能會使得一些點(diǎn)又變成了度數(shù)小于2的

題解：

用類似拓?fù)渑判虻乃枷肴プ鼍蚾k啦

代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>using namespace std;const int N=200100;
long long a[N],ans;
int n,m,T,cnt,ok[N],vis[N],pre[N],nxt[N],to[N],tot[N],col;
vector<int> s[N];
queue<int> q;void dfs(int x,int fa)
{s[col].push_back(x);ok[x]=0;for(int p=pre[x];p!=-1;p=nxt[p]){if((!vis[p])||(!ok[to[p]])) continue;if(p==(fa^1)) continue;dfs(to[p],p);}
}
void makeedge(int x,int y)
{to[cnt]=y;nxt[cnt]=pre[x];pre[x]=cnt++;to[cnt]=x;nxt[cnt]=pre[y];pre[y]=cnt++;
}int main()
{scanf("%d",&T);while(T--){memset(tot,0,sizeof(tot));memset(pre,-1,sizeof(pre));memset(ok,1,sizeof(ok));memset(vis,1,sizeof(vis));ans=0LL;cnt=0;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%I64d",&a[i]);}for(int i=0;i<m;i++){int x,y;scanf("%d%d",&x,&y);tot[x]++;tot[y]++;makeedge(x,y);}while(!q.empty()) q.pop();for(int i=1;i<=n;i++){if(tot[i]<2){q.push(i);}}while(!q.empty()){int x=q.front();q.pop();ok[x]=0;for(int p=pre[x];p!=-1;p=nxt[p]){vis[p]=0;tot[x]--;tot[to[p]]--;if(ok[to[p]]&&tot[to[p]]<2){q.push(to[p]);}}}col=0;for(int i=1;i<=n;i++){col++;s[col].clear();if(ok[i]){dfs(i,cnt+10);if(s[col].size()%2==1){for(int j=0;j<s[col].size();j++){ans+=a[s[col][j]];}}}}printf("%I64d\n",ans);}return 0;
}

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轉(zhuǎn)載于:https://www.cnblogs.com/qscqesze/p/4805256.html

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