Pseudoforest |
| Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) |
| Total Submission(s): 389 Accepted Submission(s): 165 |
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Problem Description In graph theory, a pseudoforest is an undirected graph in which every connected component has at most one cycle. The maximal pseudoforests of G are the pseudoforest subgraphs of G that are not contained within any larger pseudoforest of G. A pesudoforest is larger than another if and only if the total value of the edges is greater than another one’s. |
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Input The input consists of multiple test cases. The first line of each test case contains two integers, n(0 < n <= 10000), m(0 <= m <= 100000), which are the number of the vertexes and the number of the edges. The next m lines, each line consists of three integers, u, v, c, which means there is an edge with value c (0 < c <= 10000) between u and v. You can assume that there are no loop and no multiple edges. |
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Output Output the sum of the value of the edges of the maximum pesudoforest. |
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Sample Input 3 3 0 1 1 1 2 1 2 0 1 4 5 0 1 1 1 2 1 2 3 1 3 0 1 0 2 2 0 0 |
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Sample Output 3 5 |
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Source “光庭杯”第五届华中北区程序设计邀请赛 暨 WHU第八届程序设计竞赛 |
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Recommend lcy |
/*
初级想方法，最大生成树再加一条最长边
讲解：没看明白题意的傻逼想法，题目说不是最大生成树，森林也可以，但是最多只能有一个环
正解：将所有的边都加到树上，加的时候如果两点在同一个并查集：如果原来有环就不能加
如果不在同一个并查集：如果原来两个都有环不能加
*/
#include<bits/stdc++.h>
using namespace std;
struct node
{
int u,v,val;
node()
{}
node(int a,int b,int c)
{
u=a;
v=b;
val=c;
}
bool operator < (const node &a) const
{
return val>a.val;
}
};
vector<node>edge;
int bin[10005];
int n,m;
int x,y,val;
int h[10005];//表示当前集合有没有环
long long cur=0;
void init()
{
for(int i=0;i<=n;i++)
{
bin[i]=i;
h[i]=0;
}
edge.clear();
cur=0;
}
int findx(int x)
{
int temp=x;
while(x!=bin[x])
x=bin[x];
bin[temp]=x;
return x;
}
int main()
{
//freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
{
init();
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&x,&y,&val);
edge.push_back(node(x,y,val));
}
sort(edge.begin(),edge.end());
for(int i=0;i<edge.size();i++)
{
int fx=findx(edge[i].u);
int fy=findx(edge[i].v);
if(fx==fy)//两个原来就是一个并查集的就可能产生环了
{
if(h[fx])//有环了
continue;
cur+=edge[i].val;
h[fx]=1;
}
else
{
if(h[fx]&&h[fy])//两个集合都有环不可以
continue;
bin[fy]=fx;
cur+=edge[i].val;
if(h[fx]||h[fy])
h[fx]=1;
}
}
printf("%lld\n",cur);
}
return 0;
}