HDU-3367 最大生成树

最新推荐文章于 2019-04-10 21:40:46 发布

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最新推荐文章于 2019-04-10 21:40:46 发布

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分类专栏： hdu ACM_树图

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本文探讨了图论中的伪森林概念，即每个连通组件最多包含一个环的无向图。通过Kruskal算法解决寻找具有最大边权总和的伪森林子图问题，确保每个连通组件的环数限制。

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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.
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.
The last test case is followed by a line containing two zeros, which means the end of the input.
Output
Output the sum of the value of the edges of the maximum pesudoforest.

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
Sample Output
3
5
题意：
求伪森林的最大的权值和；并且每个连通分量中最多只能有一个环；
解题思路：
kruskal 算法
代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=10005;
int pre[maxn],cir[maxn];
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
int find(int x){return -1==pre[x]?x:pre[x]=find(pre[x]);}
struct poi{
	int u,v,c;
	bool operator <(const poi &t)const{
		return c<t.c;//最大生成树
	}
};
priority_queue<poi>q;
void kru(){
	int ans=0;
	while(!q.empty()){
		poi now=q.top();q.pop();
		now.u=find(now.u),now.v=find(now.v);
		if(now.u!=now.v){
			if(cir[now.u]&&cir[now.v])continue;//如果都有环
			cir[now.v]=cir[now.u]||cir[now.v];//有环？
			pre[now.u]=now.v;
			ans+=now.c;
		}
		else if(!cir[now.u])  cir[now.u]=1,ans+=now.c;//有环
	}
	printf("%d\n",ans);
}
int main(){
	while(1){
		int n=read(),m=read();
		if(!n&&!m)break;
		memset(pre,-1,sizeof(pre));
		memset(cir,0,sizeof(cir));
		while(m--)q.push({read(),read(),read()});
		kru();
	}
	return 0;
}