[ZJOI2009]BZOJ1412 狼和羊的故事-最小割

传送门

题目大意：自行参考

题解：

SB最小割

任意两个相邻格子之间连边容量为1，a[i][j]=1的(i,j)和s相连，=2和t相连，这两个容量都是正无穷。

跑最小割即可。

//BZOJ 1412 //ZJOI 2009 #include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<queue>#include<climits>#define MAXN 11000#define MAXM 160000#define INF INT_MAX#define build_edge(u,v,c) add_edge(u,v,c),add_edge(v,u,0)#define P(x,y) (~-(x)*m+(y))#define inrange(x,y) (1<=(x)&&(x)<=n&&1<=(y)&&(y)<=m)using namespace std;struct edges{int to,pre,resf;}e[MAXM];int h[MAXN],a[110][110],cur[MAXN],lev[MAXN];bool vis[MAXN];queue<int> q;int etop;int dx[]={-1,0,1,0},dy[]={0,1,0,-1};int add_edge(int u,int v,int c){etop++;e[etop].to=v;e[etop].resf=c;e[etop].pre=h[u];h[u]=etop;return 0;}bool bfs(int s,int t){memset(vis,false,sizeof(vis));memset(lev,0,sizeof(lev));while(!q.empty()) q.pop();q.push(s);lev[s]=0;vis[s]=true;while(!q.empty()){int x=q.front();q.pop();for(int i=h[x];i;i=e[i].pre)if(!vis[e[i].to]&&e[i].resf){vis[e[i].to]=true;lev[e[i].to]=lev[x]+1;q.push(e[i].to);}}return vis[t];}int dfs(int s,int t,int a){if(s==t||!a) return a;int flow=0,f;for(int &i=cur[s];i;i=e[i].pre)if(lev[e[i].to]==lev[s]+1&&(f=dfs(e[i].to,t,min(e[i].resf,a)))>0){a-=f;flow+=f;e[i].resf-=f;e[((i-1)^1)+1].resf+=f;if(!a) break;}return flow;}int main(){int n,m;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)scanf("%d",&a[i][j]);int s=0,t=n*m+1;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(a[i][j]==1) build_edge(s,P(i,j),INF);if(a[i][j]==2) build_edge(P(i,j),t,INF);for(int k=0;k<4;k++){int x=i+dx[k],y=j+dy[k];if(inrange(x,y)) build_edge(P(i,j),P(x,y),1);}}int flow=0;while(bfs(s,t)){for(int i=s;i<=t;i++)cur[i]=h[i];flow+=dfs(s,t,INF);}printf("%d\n",flow);return 0;}