hdu3046 最小割

Pleasant sheep and big big wolf

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1601    Accepted Submission(s): 687

Problem Description

In ZJNU, there is a well-known prairie. And it attracts pleasant sheep and his companions to have a holiday. Big big wolf and his families know about this, and quietly hid in the big lawn. As ZJNU ACM/ICPC team, we have an obligation to protect pleasant sheep and his companions to free from being disturbed by big big wolf. We decided to build a number of unit fence whose length is 1. Any wolf and sheep can not cross the fence. Of course, one grid can only contain an animal.
Now, we ask to place the minimum fences to let pleasant sheep and his Companions to free from being disturbed by big big wolf and his companions. 

 

Input

There are many cases. 
For every case: 

N and M（N,M<=200）
then N*M matrix: 
0 is empty, and 1 is pleasant sheep and his companions, 2 is big big wolf and his companions.

 

Output

For every case:

First line output “Case p:”, p is the p-th case; 
The second line is the answer. 

 

Sample Input

   
   
   
   

    
    
    
    4 6
1 0 0 1 0 0
0 1 1 0 0 0
2 0 0 0 0 0
0 2 0 1 1 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1:
4
   
   
   
   

 

Source

2009 Multi-University Training Contest 14 - Host by ZJNU

 

Recommend

gaojie

 

#include <iostream>
#include <cstdio>
#include <cstring>
#define inf 99999999
using namespace std;
int map[205][205];
struct node
{
    int u,v,f;
};
node e[1000000];
int first[50000],next[1000000],cc;
int d[50000],gap[50000],curedge[50000],pre[50000];
int xx[4]={0,1,0,-1};
int yy[4]={1,0,-1,0};
inline void add_edge(int u,int v,int f)
{
    e[cc].u=u;
    e[cc].v=v;
    e[cc].f=f;
    next[cc]=first[u];
    first[u]=cc;
    cc++;

    e[cc].u=v;
    e[cc].v=u;
    e[cc].f=0;
    next[cc]=first[v];
    first[v]=cc;
    cc++;
};
int sap_max_flow(int s,int t,int n)
{
    int cur_flow=0,flow_ans=0,i,u,neck,tmp;
    memset(d,0,sizeof(d));
    memset(gap,0,sizeof(gap));
    memset(pre,-1,sizeof(pre));
    for(i=0;i<=n;i++)
        curedge[i]=first[i];
    gap[0]=n+1;
    u=s;
    while(d[s]<n+1)
    {

        if(u==t)
        {
            cur_flow=inf;
            for(i=s;i!=t;i=e[curedge[i]].v)
            {
                if(cur_flow>e[curedge[i]].f)
                    cur_flow=e[curedge[i]].f,neck=i;
            }
            for(i=s;i!=t;i=e[curedge[i]].v)
            {
                tmp=curedge[i];
                e[tmp].f-=cur_flow;
                e[tmp^1].f+=cur_flow;
            }
            flow_ans+=cur_flow;
            u=neck;
        }
        for(i=curedge[u];i!=-1;i=next[i])
            if(e[i].f&&d[u]==d[e[i].v]+1)
                break;

        if(i!=-1)
        {
            curedge[u]=i;
            pre[e[i].v]=u;
            u=e[i].v;
        }
        else
        {
            if(0==--gap[d[u]])
                break;
            curedge[u]=first[u];
            for(tmp=n+5,i=first[u];i!=-1;i=next[i])
                if(e[i].f)
                    tmp=min(tmp,d[e[i].v]);
            d[u]=tmp+1;
            ++gap[d[u]];
            if(u!=s)
                u=pre[u];
        }
    }
    return flow_ans;

}
int main()
{
    int n,m;
    int tt=0;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int i,j;
        memset(map,0,sizeof(map));
        memset(e,0,sizeof(e));
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
                scanf("%d",&map[i][j]);
        cc=0;
        int s=n*m;
        int t=n*m+1;
        memset(first,-1,sizeof(first));
        memset(next,-1,sizeof(next));
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                int k;
                for(k=0;k<4;k++)
                {
                    int tx=i+xx[k];
                    int ty=j+yy[k];
                    if(tx<0||tx>=n||ty<0||ty>=m)
                        continue;
                    add_edge(i*m+j,tx*m+ty,1);
                }
                if(map[i][j]==1)
                    add_edge(i*m+j,t,inf);
                if(map[i][j]==2)
                    add_edge(s,i*m+j,inf);
            }
        }
        printf("Case %d:\n",++tt);
        printf("%d\n",sap_max_flow(s,t,t));
    }
    return 0;
}