[POJ2125 Destroying The Graph]

[关键字]：图论 网络流

[题目大意]：给出一个有向图，定义两种操作：w+(i)删掉i点的所有入边w-(i)删掉i点的所有出边，每种操作都有对应的花费，问删掉所有边的最小花费。

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[分析]：详细的解法还是看《最小割模型在信息学竞赛中的应用》这篇论文吧。

[代码]：

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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN=11000;
const int MAXL=209;
const int INF=0x7fffffff;

struct node
{
       int x,y,d,next,op;
}e[MAXN];
int n,m,tot,S,T,n2;
int first[MAXL],rec[MAXL];
int num[MAXL],h[MAXL],v[MAXL],g[MAXL];

void Add(int x,int y,int d)
{
     ++tot;
     e[tot].x=x;
     e[tot].y=y;
     e[tot].d=d;
     e[tot].next=first[x];
     e[tot].op=tot+1;
     first[x]=tot;
     ++tot;
     e[tot].x=y;
     e[tot].y=x;
     e[tot].d=0;
     e[tot].next=first[y];
     e[tot].op=tot-1;
     first[y]=tot;
}

int Find(int u,int flow)
{
    if (u==T) return flow;
    int pos=n-1,temp=flow;
    for (int i=first[u];i;i=e[i].next)
    {
        if (h[u]==h[e[i].y]+1 && e[i].d>0)
        {
                              int f=Find(e[i].y,min(temp,e[i].d));
                              temp-=f;
                              e[i].d-=f;
                              e[e[i].op].d+=f;
                              if (h[S]==n || !temp) return flow-temp;
        }
        if (e[i].d>0 && pos>h[e[i].y]) pos=h[e[i].y];
    }
    if (flow==temp)
    {
                   --num[h[u]];
                   if (num[h[u]]==0) h[S]=n;
                   else
                   {
                       h[u]=pos+1;
                       ++num[h[u]];
                   }
    }
    return flow-temp;
}

void DFS(int u)
{
     v[u]=0;
     g[u]=1;
     for (int i=first[u];i;i=e[i].next)
         if (!g[e[i].y] && e[i].d>0) DFS(e[i].y);
}

void Solve()
{
     memset(num,0,sizeof(num));
     memset(h,0,sizeof(h));
     num[0]=n;
     int ans=0;
     while (h[S]<n) ans+=Find(S,INF);
     printf("%d\n",ans);     
     for (int i=S;i<=T;++i) v[i]=1;
     memset(g,0,sizeof(g));
     int t=0;
     DFS(S);
     for (int i=1;i<=tot;i+=2)
         if (v[e[i].x]==0 && v[e[i].y]==1 && e[i].d==0) rec[++t]=i;
     printf("%d\n",t);
     for (int i=1;i<=t;++i)
     {
         if (e[rec[i]].x==S)printf("%d -\n",e[rec[i]].y); else printf("%d +\n",e[rec[i]].x-n2);
         //printf("%d %d\n",e[rec[i]].x,e[rec[i]].y);
     }
}

int main()
{
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    scanf("%d%d",&n,&m);
    S=0,T=n+n+1;
    int x,y;
    for (int i=1;i<=n;++i)
    {
        scanf("%d",&x);
        Add(n+i,T,x);
    }
    for (int i=1;i<=n;++i)
    {
        scanf("%d",&x);
        Add(S,i,x);
    }
    for (int i=1;i<=m;++i)
    {
        scanf("%d%d",&x,&y);
        Add(x,y+n,INF);
    }
    n2=n;
    n=n*2+2;
    Solve();
    //printf("%d %d\n",n,n2);
    return 0;
}