求全局最小割（SW算法）

hdu3002

King of Destruction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1022    Accepted Submission(s): 400

Problem Description

Zhou xingxing is the successor of one style of kung fu called "Karate Kid".he is falling love with a beautiful judo student,after being humiliated by her boyfriend,a Taekwando master from Japan,Zhou is going to fight with his rival in love.The way they fight is to destroy the wooden plank between some wooden pegs,in order to cut these wooden pegs into two disconnected parts,and destroy each piece of plank need consume different energy.However Zhou xingxing is beginner after all,so he is turn to you for help,please calculate the minimum energy he need to destroy the wooden plank.

 

Input

The input consists of multiple test cases.
Each test case starts with two integers n (0 < n <= 100) and m in one line, where n、m are the number of wooden pegs and wooden plank. 
Following are m lines, each line contains three integers s, e and q (0 <= s, e < n,q > 0), meaning that there need q energy to destroy the wooden plank between s and e.

 

Output

There is only one line for each test case, which contains the minimum energy they need to complete this fight.

 

Sample Input

2 10 1 503 20 1 501 2 10

 

Sample Output

5010

题意：求无向图的最小割。
思路：无向图的最小割又叫全局最小割，枚举汇点求最大流效率很低，因而普遍使用StoerWagner算法，时间复杂度为O（n^3）。
程序：
#include"stdio.h"#include"string.h"#define inf 99999999int G[666][666],ans;int min(int a,int b){    return a<b?a:b;}int f[666];int finde(int x){    if(x!=f[x])        f[x]=finde(f[x]);    return f[x];}void make(int a,int b){    int x=finde(a);    int y=finde(b);    if(x!=y)        f[x]=y;}void Mincut(int n){    ans=inf;    int node[666],dis[666];    bool use[666];    int i,k,pre;    for(i=0;i<n;i++)        node[i]=i;    while(n>1)    {        int maxj=1;        for(i=1;i<n;i++)//初始化到已圈集合的割大小        {            dis[node[i]]=G[node[i]][node[0]];            if(dis[node[maxj]]<dis[node[i]])            {                maxj=i;            }        }        pre=0;        memset(use,false,sizeof(use));        use[node[0]]=true;        for(k=1;k<n;k++)        {            if(k==n-1)//只剩最后一个没加入集合的点，更新最小割            {                ans=min(ans,dis[node[maxj]]);                for(i=0;i<n;i++)//合并最后一个点以及推出它的集合中的点                {                    G[node[i]][node[pre]]=G[node[pre]][node[i]]+=G[node[i]][node[maxj]];                }                node[maxj]=node[--n];//缩点后的图            }            use[node[maxj]]=true;            pre=maxj;            maxj=-1;            for(i=1;i<n;i++)            {                if(!use[node[i]])                {                    dis[node[i]]+=G[node[i]][node[pre]];//将上次求的maxj加入集合，合并与它相邻的边到割集                    if(maxj==-1||dis[node[maxj]]<dis[node[i]])                        maxj=i;                }            }        }    }}int main(){    int n,m,i;    while(scanf("%d%d",&n,&m)!=-1)    {        memset(G,0,sizeof(G));        for(i=0;i<n;i++)        {            f[i]=i;        }        while(m--)        {            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            G[a][b]=G[b][a]+=c;            make(a,b);        }        int flag=0;        for(i=0;i<n;i++)        {            if(finde(i)!=finde(0))                flag++;        }        if(flag)            printf("0\n");        else        {            Mincut(n);            printf("%d\n",ans);        }    }}