Pleasant sheep and big big wolf

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

题意：在一个n*m的草原上，羊是1，狼是2，要安长度为1的栅栏，要求用最小的栅栏保证羊不被狼吃掉。

思路：建立一个超级源点与狼相连，长度为inf，建立一个超级汇点与羊相连，长度为inf，然后剩下的格子相连长度为1，然后找。

这个图真心怪怪的，一开始用搜索找四个方向，愣是不对，最后只好用笨办法，一个一个来喽。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
#include <map>

using namespace std;
const int inf=0x3f3f3f3f;
int head[1000100],num[100010],d[100010],pre[100010],cur[100010],mp[310][310];
int n,m,cnt,nv,s,t;
int jx[]={1,0,-1,0};
int jy[]={0,1,0,-1};
struct node
{
    int u,v,cap;
    int next;
}edge[10000010];

void add(int u, int v, int cap)
{
    edge[cnt].v=v;
    edge[cnt].cap=cap;
    edge[cnt].next=head[u];
    head[u]=cnt++;

    edge[cnt].v=u;
    edge[cnt].cap=0;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}

void bfs()
{
    memset(d,-1,sizeof(d));
    memset(num,0,sizeof(num));
    queue<int >q;
    q.push(t);
    d[t]=0;
    num[0]=1;
    while(!q.empty()) {
        int i;
        int u=q.front();
        q.pop();
        for(i=head[u]; i!=-1; i=edge[i].next) {
            int v=edge[i].v;
            if(d[v]==-1) continue;
            d[v]=d[u]+1;
            num[d[v]]++;
            q.push(v);
        }
    }
}

void isap()
{
    memcpy(cur,head,sizeof(cur));
    int flow=0, u=pre[s]=s, i;
    bfs();
    while(d[s]<nv) {
        if(u==t) {
            int f=inf, pos;
            for(i=s; i!=t; i=edge[cur[i]].v) {
                if(f>edge[cur[i]].cap) {
                    f=edge[cur[i]].cap;
                    pos=i;
                }
            }
            for(i=s; i!=t; i=edge[cur[i]].v) {
                edge[cur[i]].cap-=f;
                edge[cur[i]^1].cap+=f;
            }
            flow+=f;
            u=pos;
        }
        for(i=cur[u]; i!=-1; i=edge[i].next) {
            if(d[edge[i].v]+1==d[u]&&edge[i].cap)
                break;
        }
        if(i!=-1) {
            cur[u]=i;
            pre[edge[i].v]=u;
            u=edge[i].v;
        } else {
            if(--num[d[u]]==0) break;
            int mind=nv;
            for(i=head[u]; i!=-1; i=edge[i].next) {
                if(mind>d[edge[i].v]&&edge[i].cap) {
                    mind=d[edge[i].v];
                    cur[u]=i;
                }
            }
            d[u]=mind+1;
            num[d[u]]++;
            u=pre[u];
        }
    }
    printf("%d\n",flow);
}

int main()
{
    int i, j,k;
    int tt=1;
    while(~scanf("%d%d",&n,&m))
    {
        cnt=0;
        s=0;
        t=n*m+1;
        nv=t+1;
        memset(head,-1,sizeof(head));
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
            {
                scanf("%d",&mp[i][j]);
                if(mp[i][j]==1)
                    add(s,(i-1)*m+j,inf);
                if(mp[i][j]==2)
                    add((i-1)*m+j,t,inf);
                if(i!=1)
                    add((i-1)*m+j,(i-2)*m+j,1);// 与上面的点相连
                if(i!=n)
                    add((i-1)*m+j,i*m+j,1);//与下面的点相连
                if(j!=1)
                    add((i-1)*m+j,(i-1)*m+j-1,1);//与左边的点相连
                if(j!=m)
                    add((i-1)*m+j,(i-1)*m+j+1,1);//与右边的点相连
                }
            }
        printf("Case %d:\n",tt++);
        isap();
    }
    return 0;
}