【ARC074F】Lotus Leaves 【最小割】

最小割的裸题。先拆点。之后暴力边可过，但更好的方法是每行每列建虚点，每片莲叶像所在行列互相连边。然后跑最大流就好。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
using namespace std;
const int inf=0x3f3f3f3f;
int n,m,bx,by,ex,ey,s,t,cnt,tot,id[105][105],head[100005],dep[100005],to[1000005],nxt[1000005],dd[1000005];
char str[105][105];
queue<int> q;
void adde(int u,int v,int d){
    to[++cnt]=v;
    nxt[cnt]=head[u];
    dd[cnt]=d;
    head[u]=cnt;
    to[++cnt]=u;
    nxt[cnt]=head[v];
    dd[cnt]=0;
    head[v]=cnt;
}
bool bfs(){
    for(int i=s;i<=t;i++){
        dep[i]=0;
    }
    dep[s]=1;
    q.push(s);
    while(!q.empty()){
        int u=q.front(),v;
        q.pop();
        for(int i=head[u];i;i=nxt[i]){
            v=to[i];
            if(dd[i]&&!dep[v]){
                dep[v]=dep[u]+1;
                q.push(v);
            }
        }
    }
    return dep[t];
}
int dfs(int u,int f){
    if(u==t||!f){
        return f;
    }
    int v,tmp,flow=0;
    for(int i=head[u];i&&f;i=nxt[i]){
        v=to[i];
        if(dd[i]&&dep[v]==dep[u]+1&&(tmp=dfs(v,min(f,dd[i])))){
            dd[i]-=tmp;
            dd[i^1]+=tmp;
            f-=tmp;
            flow+=tmp;
        }
    }
    if(!flow){
        dep[u]=0;
    }
    return flow;
}
int maxflow(){
    int flow=0;
    while(bfs()){
        flow+=dfs(s,inf);
    }
    if(flow>=inf){
        puts("-1");
        exit(0);
    }
    return flow;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%s",str[i]+1);
        for(int j=1;j<=m;j++){
            if(str[i][j]=='S'){
                bx=i;
                by=j;
            }else if(str[i][j]=='T'){
                ex=i;
                ey=j;
            }else if(str[i][j]=='o'){
                tot++;
                id[i][j]=n+m+tot;
            }
        }
    }
    if(bx==ex||by==ey){
        puts("-1");
        return 0;
    }
    s=0,t=n+m+2*tot+1,cnt=1;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(str[i][j]=='o'){
                adde(id[i][j],id[i][j]+tot,1);
                adde(id[i][j]+tot,i,inf);
                adde(i,id[i][j],inf);
                adde(id[i][j]+tot,j+n,inf);
                adde(j+n,id[i][j],inf);
            }
        }
    }
    adde(s,bx,inf);
    adde(s,by+n,inf);
    adde(ex,t,inf);
    adde(ey+n,t,inf);
    printf("%d
",maxflow());
    return 0;
}