网络流模板 POJ 1273 Drainage Ditches

来源：互联网发布：linux 查看进程命令行编辑：程序博客网时间：2024/06/06 00:30
邻接表的ISAP:
#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <stack>#include <map>#include <limits.h>using namespace std;const int inf=1e8-1;const int N=1e5+5;int n,m;struct node{    int u,v;    int flow;} e[N];int first[1000];int next[N];int k,nn;int isap(int s,int t){    int num[1000],d[1000],cure[1000],pre[1000];    int cur,ans=0,u,tmp,neck;    memset(d,0,sizeof(d));    memset(num,0,sizeof(num));    memset(pre,-1,sizeof(pre));    for(int i=0; i<=nn; i++)        cure[i]=first[i];    num[0]=nn+1;    u=s;    while(d[t]<nn+1)    {        if(u==t)        {            cur=inf;            for(int i=s; i!=t; i=e[cure[i]].v)                if(cur>e[cure[i]].flow)                {                    neck=i;                    cur=e[cure[i]].flow;                }            for(int i=s; i!=t; i=e[cure[i]].v)            {                tmp=cure[i];                e[tmp].flow-=cur;                e[tmp^1].flow+=cur;            }            ans+=cur;            u=neck;        }        int i;        for(i=cure[u]; i!=-1; i=next[i])            if(e[i].flow&&d[u]==d[e[i].v]+1)                break;        if(i!=-1)        {            cure[u]=i;            pre[e[i].v]=u;            u=e[i].v;        }        else        {            if(0==--num[d[u]])                break;            cure[u]=first[u];            for(tmp=nn+1,i=first[u]; i!=-1; i=next[i])                if(e[i].flow)                    tmp=min(tmp,d[e[i].v]);            d[u]=tmp+1;            ++num[d[u]];            if(u!=s)                u=pre[u];        }    }    return ans;}inline void add(int u,int v,int w){   // printf("--%d %d %d\n",u,v,w);    e[k].u=u;    e[k].v=v;    e[k].flow=w;    next[k]=first[u];    first[u]=k++;    e[k].u=v;    e[k].v=u;    e[k].flow=0;    next[k]=first[v];    first[v]=k++;}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(first,-1,sizeof(first));        memset(next,-1,sizeof(next));        k=0;        int a,b,c;        for(int i=1; i<=n; i++)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,c);        }        nn=m;        printf("%d\n",isap(1,nn));    }}
0 0