Ponds

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

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

先把边存起来，度计算出来，按照拓扑序删点(单点要注意，这里wa了一次)，然后扫描一遍边，有删掉的点就忽略，用并查集维护连通分量的结点数和总权值。

#include<bits/stdc++.h>

using namespace std;

const int maxn = 1e4+,maxm = 2e5+;

int n,m;

int val[maxn];

int head[maxn],nxt[maxm],to[maxm];

int deg[maxn],ecnt;

bool rmvd[maxn];

void addEdge(int u,int v)

{

    to[ecnt] = v;

    nxt[ecnt] = head[u];

    head[u] = ecnt++;

    deg[u]++;

}

void topo()

{

    queue<int> q;

    for(int i = ; i <= n; i++){

        if(deg[i] <= ){

            rmvd[i] = true;

            q.push(i);

        }

    }

    while(q.size()){

        int u = q.front(); q.pop();

        for(int i = head[u]; ~i; i = nxt[i]){

            int v = to[i];

            if(!rmvd[v] && --deg[v] == ){

                q.push(v); rmvd[v] = true;

            }

        }

    }

}

long long sum[maxn];

int pa[maxn],cnt[maxn];

int fdst(int x) { return x==pa[x]?x:pa[x]=fdst(pa[x]); }

int main()

{

    //freopen("in.txt","r",stdin);

    int T; scanf("%d",&T);

    while(T--){

        scanf("%d%d",&n,&m);

        for(int i = ; i <= n; i++) scanf("%d",val+i);

        memset(head,-,sizeof(head));

        memset(deg,,sizeof(deg));

        memset(rmvd,,sizeof(rmvd));

        ecnt = ;

        for(int i = ; i < m; i++){

            int u,v; scanf("%d%d",&u,&v);

            addEdge(u,v); addEdge(v,u);

        }

        topo();

        for(int i = ; i <= n; i++) pa[i] = i,sum[i] = val[i],cnt[i] = ;

        for(int i = ,M = *m; i < M; i += ){

            int u = to[i], v = to[i^];

            if(!rmvd[u] && !rmvd[v]){

                int a = fdst(u), b = fdst(v);

                if(a != b){

                    pa[a] = b;

                    sum[b] += sum[a];

                    cnt[b] += cnt[a];

                }

            }

        }

        long long ans = ;

        for(int i = ; i <= n ;i++){

            int f = fdst(i);

            if(!rmvd[f]){

                if(cnt[f]&){

                    ans += sum[f];

                }

                rmvd[f] = true;

            }

        }

        printf("%I64d\n",ans);

    }

    return ;

}

巴特西

2015 ACM/ICPC Asia Regional Changchun Online Pro 1002 Ponds（拓扑排序+并查集）

Ponds

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