HDU3605 Escape(最大流)

2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets. 

Input

More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet. 
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most.. 
0 <= ai <= 100000 

Output

Determine whether all people can live up to these stars 
If you can output YES, otherwise output NO. 

Sample Input

1 1
1
1

2 2
1 0
1 0
1 1

Sample Output

YES
NO

最大流

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod=609929123;
const int N=100220;
int frist[N];
int current[N];
int dis[N];
char mapp[100][100];
char mapp1[100][100];
int mp[100][100];
int number[100][100];
int R,S,T,t,n,d,m,cnt;
struct node
{
    int v,cap,next;//领接表
}es[4*N];
int addedge(int u,int v,int cap)
{
    node  e1={v,cap,frist[u]};
    es[R]=e1;
    frist[u]=R++;
    node e2={u,0,frist[v]};
    es[R]=e2;
    frist[v]=R++;
}
int bfs()
{
  queue<int>q;
  q.push(S);
  memset(dis,-1,sizeof(dis));
  dis[S]=0;
  while(!q.empty())
  {
      int h=q.front();
      if(h==T) return 1;
      q.pop();
      for(int i=frist[h];i!=-1;i=es[i].next)
      {
          int temp=es[i].v;
          if(dis[temp]==-1&&es[i].cap)
          {
              dis[temp]=dis[h]+1;
              q.push(temp);
          }
      }
  }
  return 0;
}
int dinic(int x,int maxflow)
{
  if(x==T) return maxflow;
  int flow,f=0;
  for(int &i=current[x];i!=-1;i=es[i].next)
  {
      int temp=es[i].v;
      if(dis[temp]==dis[x]+1&&es[i].cap)
      {
      flow=dinic(temp,min(maxflow-f,es[i].cap));
      es[i].cap-=flow;
      es[i^1].cap+=flow;
      f+=flow;
      if(f==maxflow) break;
      }
  }
  return f;
}
int DINIC()
{
 int ans=0;
 while(bfs())
 {
     int flow;
     memcpy(current,&frist,sizeof(frist));//复制
     while(flow=dinic(S,inf))
     ans+=flow;
 }
 return ans;
}
int main()
{
   scanf("%d",&t);
    int o=0;
   while(t--)
   {
       memset(frist,-1,sizeof(frist));
       memset(mapp,0,sizeof(mapp));
        memset(mapp1,0,sizeof(mapp1));
        memset(mp,0,sizeof(mp));
        memset(number,0,sizeof(number));
       R=0,S=0;
        cnt=0;
       scanf("%d%d",&n,&d);
       for(int i=1;i<=n;i++)
       {
           scanf("%s",mapp[i]+1);
           m=strlen(mapp[i]+1);
           for(int j=1;j<=m;j++)
           {
               mp[i][j]=mapp[i][j]-'0';
               number[i][j]=++cnt;
           }
       }
        T=2*cnt+1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(mp[i][j])
                {
                    if(i<=d||j<=d||i+d>n||j+d>m)
                        addedge(number[i][j]+cnt,T,inf);
                    addedge(number[i][j],number[i][j]+cnt,mp[i][j]);
                    for(int k=1;k<=n;k++)
                    {
                        for(int p=1;p<=m;p++)
                        {
                            int dx=abs(i-k);
                            int dy=abs(j-p);
                            double mm=sqrt(dx*dx*1.0+dy*dy*1.0);
                            if(mm>d) continue;
                            addedge(number[i][j]+cnt,number[k][p],inf);
                        }
                    }
                }
            }
        }
        int ll=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%s",mapp1[i]+1);
            for(int j=1;j<=m;j++)
            {
                if(mapp1[i][j]=='L')
                {
                    ll++;
                    addedge(0,number[i][j],1);
                }
            }
        }
        ll-=DINIC();
        if(ll==0)
        printf("Case #%d: no lizard was left behind.\n",++o);
        else if(ll==1)
            printf("Case #%d: 1 lizard was left behind.\n",++o);
        else
       printf("Case #%d: %d lizards were left behind.\n",++o,ll);

   }
}