hdu-3367 Pseudoforest（伪森林， kruskal变形）

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Pseudoforest

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2401    Accepted Submission(s): 943

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 30 1 11 2 12 0 14 50 1 11 2 12 3 13 0 10 2 20 0

 

Sample Output

35

题意：每一个联通子集顶多只能有一个环，在给定边里求满足条件的最大总长度

思路：和最大生成树很像。。不过这题是可以分成多个联通子集的，注意处理下最多只能有一个环就可以了

两个联通子集，都有环的时候肯定不能连接了， 不然新的联通子集就有两个环了。

其中一个有环的时候可以连接，注意标记新的联通子集有环

都没环的时候当然可以连接，且新的联通子集无环。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 10010#define M 100100int n,m;int pre[N];int flag[N];struct Edge{    int u,v,w;} e[M];void init(){    for(int i=0; i<n; i++)        pre[i]=i;    memset(flag,0,sizeof(flag));}bool cmp(Edge a,Edge b){    return a.w>b.w;}int finds(int x){    int r=x;    while(r!=pre[r])        r=pre[r];    int j;    while(x!=pre[x])    {        j=x;        x=pre[x];        pre[j]=r;    }    return r;}int kruskal(){    int ans=0,x,y;    for(int i=0; i<m; i++)    {        x=finds(e[i].u);        y=finds(e[i].v);        if(x==y)        {            if(!flag[x])            {                flag[x]=1;                ans+=e[i].w;            }            continue;        }        if(flag[x]&&flag[y]) continue;        pre[x]=y;        ans+=e[i].w;        if(!flag[x]&&!flag[y]) continue;        flag[x]=flag[y]=1;    }    return ans;}int main(){    int u,v,w;    while(scanf("%d %d",&n,&m)&&n)    {        init();        for(int i=0; i<m; i++)            scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].w);        sort(e,e+m,cmp);        printf("%d\n",kruskal());    }    return 0;}