hdu 3549 Flow Problem 增广路ford-fullkerson算法

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const int N=1024*2;
const int inf=1<<24;

struct arc
{
    int c,f;
} g[N][N];

int n,m,i,j,u,v,c,f;

int abs(int x)
{
    if(x<0) return -x;
    else return x;
}

void ford()
{
    int p[N],a[N],flag[N];
    queue<int>q;
    while(1)
    {
        memset(p,0xff,sizeof(p));
        memset(a,0xff,sizeof(a));
        memset(flag,0xff,sizeof(flag));
        while(!q.empty()) q.pop();
        q.push(0);
        flag[0]=0;
        p[0]=0;
        a[0]=inf;
        while(!q.empty()&&flag[n-1]==-1)
        {
            u=q.front();
            q.pop();
            for(i=0; i<n; i++)
            {
                if(flag[i]==-1)
                {
                    if(g[u][i].c<inf&&g[u][i].f<g[u][i].c)
                    {
                        flag[i]=0;
                        p[i]=u;
                        a[i]=min(a[u],g[u][i].c-g[u][i].f);
                        q.push(i);
                    }
                    else if(g[i][u].c<inf&&g[i][u].f>0)
                    {
                        flag[i]=0;
                        p[i]=-u;
                        a[i]=min(a[u],g[i][u].f);
                        q.push(i);
                    }
                }
            }
            flag[u]=1;
        }
        //printf("%d %d\n",flag[n-1],a[n-1]);
        if(flag[n-1]==-1||a[n-1]==0) break;
        int k1=n-1,k2=abs(p[k1]);
        int add=a[n-1];
        while(1)
        {
            if(g[k2][k1].f<inf)
                g[k2][k1].f+=add;
            else
                g[k1][k2].f-=add;
            if(k2==0) break;
            k1=k2;
            k2=abs(p[k2]);
        }
    }

    int flow=0;
    for(i=0; i<n; i++)
        if(g[0][i].f<inf)
            flow+=g[0][i].f;

    printf("%d\n",flow);
}

int main()
{
    int _;
    scanf("%d",&_);
    for(int k=1; k<=_; k++)
    {
        scanf("%d%d",&n,&m);
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
                g[i][j].c=g[i][j].f=inf;

        for(i=0; i<m; i++)
        {
            scanf("%d%d%d",&u,&v,&c);
            u--;
            v--;
            if(g[u][v].c==inf)
                g[u][v].c=c;
            else
                g[u][v].c+=c;
            g[u][v].f=0;
        }
        printf("Case %d: ",k);
        ford();
    }
    return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。http://xiang578.top/

posted @ 2015-08-11 14:57  xryz  阅读(310)  评论(0编辑  收藏  举报