POJ 1273，HDU 1532 Drainage Ditches(最大流)

http://acm.hdu.edu.cn/showproblem.php?pid=1532

使用网络流最大流中的DINIC算法解决

#include<stdio.h>
#include<string.h>
#define point_MAX 10000
#define edge_MAX 100000
#define INF_MAX 999999999
struct EDGE
{
    int to;/*指向的点*/
    int next;/*指向的下一条邻边*/
    int w;/*权值*/
}edge[edge_MAX];
int len;/*边的数量*/
int point[point_MAX];
int Vertex,Edge;
int d[point_MAX];
void init()/*初始化*/
{
    len=0;
    memset(point,0,sizeof(point));
}
int add_edge(int a,int b,int w)/*添加由a指向b的权值为w的边*/
{
    len++;
    edge[len].w=w;
    edge[len].to=b;
    edge[len].next=point[a];
    point[a]=len;
    return 0;/*无重边，插入*/
}
int bfs(int s)
{
    int q[point_MAX],front=0,rear=1,j,t,i;
    q[0]=s;
    memset(d,-1,sizeof(d));/**/
    d[s]=0;
    while(front<rear)
    {
       t=q[front++];
         for(j=point[t];j;j=edge[j].next)
         {
            if(d[edge[j].to]==-1&&edge[j].w>0)
            {
             d[edge[j].to]=d[t]+1;
           q[rear++]=edge[j].to;
        }
         }
    }
    if(d[Vertex]>=0)
       return 1;
    return 0;
}
long long min(long long a,long long b)
{
    return a<b?a:b;
}
long long dinic(int t,long long sum)
{
    int i,os,j;
    long long a;
    if(t==Vertex)
      return sum;
    os=sum;
    for(i=point[t];i&&sum;i=edge[i].next)
    {
       if(d[edge[i].to]==d[t]+1&&edge[i].w>0)
       {
           a=dinic(edge[i].to,min(sum,edge[i].w));
           edge[i].w-=a;
           for(j=point[edge[i].to];edge[j].to!=t;j=edge[j].next);
           edge[j].w+=a;
           sum-=a;
       }
    }
    return os-sum;
}
long long DINIC(int s)
{
     long long ans=0;
     while(bfs(s))
       ans+=dinic(s,INF_MAX);
     return ans;
}

int main()
{
     int i,j,x,y,w;
     while(scanf("%d%d",&Edge,&Vertex)!=EOF)
     {
        init();
        for(i=0;i<Edge;i++)
        {
           scanf("%d%d%d",&x,&y,&w);
           add_edge(x,y,w);
           add_edge(y,x,0);
        }
        printf("%lld
",DINIC(1));
     }
     return 0;
}