poj 3469 Dual Core CPU

题目链接：http://poj.org/problem?id=3469

题目大意：求一些任务在一个双核的cpu运行，但当两个需要交换数据的任务在不同的核运行时要有额外花费，求最小花费。

题目思路：别人的做法：源点连到点的容量为Ai,点到汇点的容量为Bi,对于m个关系建容量为w的双向边，这样就转化为了求图上的最小割。不过我没法准确证明。

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<string>#include<queue>#include<algorithm>#include<vector>#include<stack>#include<list>#include<iostream>#include<map>using namespace std;#define inf 0x7f3f3f3f#define Max 30000int max(int a,int b){    return a>b?a:b;}int min(int a,int b){    return a<b?a:b;}int dis[Max],gap[Max],pre[Max],cur[Max],p[Max],sum;//int d[4][2]={0,1,1,0,0,-1,-1,0};//int mp[220][220];int n,m,s,t,eid;struct node {     int to,next,c; }e[1000000];void addedge(int u,int v,int c){    e[eid].to=v;    e[eid].c=c;    e[eid].next=p[u];    p[u]=eid++;}void  ISAP(int st,int ed,int n,int count)   ///起点，终点，顶点数{    memset(dis, 0, sizeof(dis));    memset(gap, 0, sizeof(gap));  gap[0]=n;    memcpy(cur, p, sizeof(p));      ///memcpy!    int  i,flag,v,u=pre[st]=st,maxflow=0,aug=inf; //puts("akk");    while(dis[st] < n)    {        for(flag=0,i=cur[u];i!=-1; i=e[i].next)  /// cur[u]            if(e[i].c&& dis[u] == dis[e[i].to]+1)            {                flag = 1;                break;            }        if(flag)        {            if(aug > e[i].c)                aug = e[i].c;            v = e[i].to;            pre[v] = u;            cur[u] = i;            u = v;            if(u == ed)            {                for(u=pre[u]; 1;u=pre[u])    ///notice!                {                    e[cur[u]].c -= aug;                    e[cur[u]^1].c += aug;                    if(u==st) break;                   // puts("akkk");                }                maxflow += aug;                aug = inf;            }        }        else        {            int minx = n;            for(i=p[u]; i!=-1; i=e[i].next)                if(e[i].c&& dis[e[i].to]<minx)                {                    minx = dis[e[i].to];                    cur[u] = i;                }            if(--gap[dis[u]] == 0)                break;            dis[u] = minx+1;            gap[dis[u]]++;            u = pre[u];        }    }  // printf("Case %d:\n%d\n",count,maxflow);     printf("%d\n",maxflow);}int main(){    int m,n,t,count=1;    int u,v,c,i,j,k,x,y,a,b,w;    while( scanf("%d%d",&n,&m)!=EOF)    {        eid=0;        memset(p,-1,sizeof(p));        for(i=1;i<=n;i++)        {            scanf("%d%d",&a,&b);            addedge(0,i,a);            addedge(i,0,0);            addedge(i,n+1,b);            addedge(n+1,i,0);        }        while(m--)        {            scanf("%d%d%d",&a,&b,&w);            addedge(a,b,w);            addedge(a,b,0);            addedge(b,a,w);            addedge(b,a,0);        }       // printf("%d\n",eid);        ISAP(0,n+1,n+2,count++);       // printf("%d\n")    }}