題目鏈接:Lotus Leaves
顯然是一個最小割模型。
但是我們暴力連邊是O(n ^ 3)的。考慮優化:我們可以把每一條豎着的,橫着的抽出來,然後就變成O(n ^ 2)了。沒什麼難度。
AC代碼:
#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N=2e4+300,M=2e6+10;
int n,m,s,t,h[N]; char g[N][N];
int head[N],nex[M],to[M],w[M],tot=1;
inline void ade(int a,int b,int c){to[++tot]=b; nex[tot]=head[a]; w[tot]=c; head[a]=tot;}
inline void add(int a,int b,int c){ade(a,b,c); ade(b,a,0);}
inline int id(int x,int y){return (x-1)*m+y;}
inline int bfs(){
queue<int> q; q.push(s); memset(h,0,sizeof h); h[s]=1;
while(q.size()){
int u=q.front(); q.pop();
for(int i=head[u];i;i=nex[i]) if(w[i]&&!h[to[i]]) h[to[i]]=h[u]+1,q.push(to[i]);
}
return h[t];
}
int dfs(int x,int f){
if(x==t) return f; int fl=0;
for(int i=head[x];i&&f;i=nex[i]){
if(w[i]&&h[to[i]]==h[x]+1){
int mi=dfs(to[i],min(w[i],f));
w[i]-=mi,w[i^1]+=mi,fl+=mi,f-=mi;
}
}
if(!fl) h[x]=-1;
return fl;
}
inline int dinic(){
int res=0;
while(bfs()) res+=dfs(s,inf);
return res;
}
signed main(){
cin>>n>>m; t=2e4+275;
for(int i=1;i<=n;i++) scanf("%s",g[i]+1);
for(int i=1;i<=n;i++) for(int j=1;j<=m;j++){
add(id(i,j),id(i,j)+n*m,1);
if(g[i][j]=='S') add(s,id(i,j)+n*m,inf);
if(g[i][j]=='T') add(id(i,j),t,inf);
if(g[i][j]!='.'){
add(id(i,j)+n*m,n*m*2+i,inf);
add(n*m*2+i,id(i,j),inf);
add(id(i,j)+n*m,n*m*2+n+j,inf);
add(n*m*2+n+j,id(i,j),inf);
}
}
int maxflow=dinic();
if(maxflow>=0x3f3f3f3f) return puts("-1"),0;
cout<<maxflow<<endl;
return 0;
}