【链接】 我是链接,点我呀:)
【题意】

在这里输入题意

【题解】

每个点和相邻的4个点连一条容量为1的边。
然后源点和每头羊连容量INF的边
每头狼和汇点连容量INF的边。
这样求最小割的时候只会把栅栏删掉。
然后源点不能到汇点了。
显然就是每头羊都不能和狼联通了(否则肯定能有增广路

a数组一开始写成1万x1万。
(直接超时了。。

【代码】

#include <bits/stdc++.h>
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define all(x) x.begin(),x.end()
#define pb push_back
#define ms(x,y) memset(x,y,sizeof x)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;const double pi = acos(-1);
const int dx[4] = {0,0,1,-1};
const int dy[4] = {1,-1,0,0};
const int N = 10000;
const int INF = 10000+10;struct abc{int en,nex;LL flow;
};int n,m,s,t,cost[N+10],deep[N*2+20];
int fir[N+10],tfir[N+10],totm;
abc bian[2*N+2*N+N*8+10];
queue <int> dl;void add(int x,int y,LL cost){bian[totm].nex = fir[x];fir[x] = totm;bian[totm].en = y,bian[totm].flow = cost;totm++;bian[totm].nex = fir[y];fir[y] = totm;bian[totm].en = x,bian[totm].flow = 0;totm++;
}bool bfs(int s,int t){dl.push(s);ms(deep,255);deep[s] = 0;while (!dl.empty()){int x = dl.front();dl.pop();for (int temp = fir[x]; temp!= -1 ;temp = bian[temp].nex){int y = bian[temp].en;if (deep[y]==-1 && bian[temp].flow>0){deep[y] = deep[x] + 1;dl.push(y);}}}return deep[t]!=-1;
}LL dfs(int x,int t,LL limit){if (x == t) return limit;if (limit == 0) return 0;LL cur,f = 0;for (int temp = tfir[x];temp!=-1;temp = bian[temp].nex){tfir[x] = temp;int y = bian[temp].en;if (deep[y] == deep[x] + 1 && (cur = dfs(y,t,min(limit,(LL)bian[temp].flow))) ){f += cur;limit -= cur;bian[temp].flow -= cur;bian[temp^1].flow += cur;if (!limit) break;}}return f;
}
int a[100+10][100+10];int main(){#ifdef LOCAL_DEFINEfreopen("rush_in.txt", "r", stdin);#endifms(fir,255);scanf("%d%d",&n,&m);for (int i = 1;i <= n;i++)for (int j = 1;j <= m;j++)scanf("%d",&a[i][j]);for (int i = 1;i <= n;i++){for (int j = 1;j <= m;j++){for (int k = 0;k < 4;k++){int ti = i+dx[k],tj = j+dy[k];if (ti>=1 && ti<=n && tj>=1 && tj<=m){add((i-1)*m+j,(ti-1)*m+tj,1);}}}}for (int i = 1;i <= n;i++)for (int j = 1;j <=m;j++){if (a[i][j]==1){add(0,(i-1)*m+j,INF);}else if (a[i][j]==2){add((i-1)*m+j,n*m+1,INF);}}s = 0,t = n*m+1;int ans = 0;while ( bfs(s,t) ){for (int i = 0;i <= n*m+1;i++) tfir[i] = fir[i];ans += dfs(s,t,INF);}printf("%d\n",ans);return 0;
}