AtCoder Regular Contest 074F - Lotus Leaves

$n \leq 300,m \leq 300$，$n*m$的格子里有起点有终点有空地有障碍，人会从起点选一个同行或同列空地跳过去，然后一直这样跳到终点。求至少删掉多少格子使得人跳不到终点。

首先S和T同行或同列无解。

这不是裸的最小割嘛。。等会这复杂度不大对

优化：一行里的点要连来连去嘛，每个点都要向同行或同列的所有点连inf边嘛，那咱把这边聚一下，每行每列新开一个点，这一行的所有点连向代表这一行的点，代表这一行的点连向所有这一行的点，这样边从三次方变成了平方。可过。

还有一种建图方法：每行每列开一个点，把一个空地对应的行列连条边，起点向其行列连边，终点其行列向终点连边。

 //#include<iostream>
 #include<cstring>
 #include<cstdio>
 //#include<time.h>
 //#include<complex>
 //#include<queue>
 #include<algorithm>
 #include<stdlib.h>
 using namespace std;

 #define LL long long
 int qread()
 {
     char c; int s=; while ((c=getchar())<'' || c>'');
     do s=s*+c-''; while ((c=getchar())>='' && c<=''); return s;
 }

 //Pay attention to '-' and LL of qread!!!!

 int n,m;
 #define maxn 400011
 #define maxm 4000011
 struct Edge{int to,next,cap,flow;};
 struct Network
 {
     Edge edge[maxm]; int first[maxn],le,n;
     void clear(int m) {le=; n=m;}
     void in(int x,int y,int cap)
     {Edge &e=edge[le]; e.to=y; e.cap=cap; e.flow=; e.next=first[x]; first[x]=le++;}
     void insert(int x,int y,int cap) {in(x,y,cap); in(y,x,);}
     int s,t,dis[maxn],cur[maxn],que[maxn],head,tail;
     bool bfs()
     {
         memset(dis,,sizeof(dis)); dis[s]=;
         head=; tail=; que[]=s;
         while (head!=tail)
         {
             int now=que[head++];
             for (int i=first[now];i;i=edge[i].next)
             {
                 Edge &e=edge[i];
                 if (e.cap>e.flow && dis[e.to]==)
                 {
                     dis[e.to]=dis[now]+;
                     que[tail++]=e.to;
                 }
             }
         }
         return dis[t];
     }
     int dfs(int x,int a)
     {
         if (x==t || !a) return a;
         int flow=,f;
         for (int &i=cur[x];i;i=edge[i].next)
         {
             Edge &e=edge[i];
             if (dis[e.to]==dis[x]+ && (f=dfs(e.to,min(a,e.cap-e.flow)))>)
             {
                 e.flow+=f; flow+=f;
                 edge[i^].flow-=f; a-=f;
                 if (!a) break;
             }
         }
         return flow;
     }
     int Dinic(int s,int t)
     {
         this->s=s; this->t=t;
         int ans=;
         while (bfs())
         {
             for (int i=;i<=n;i++) cur[i]=first[i];
             ans+=dfs(s,0x3f3f3f3f);
         }
         return ans;
     }
 }g;

 char mp[][]; int id[][];

 int main()
 {
 //    freopen("trade.in","r",stdin);
 //    freopen("trade.out","w",stdout);
     n=qread(); m=qread();
     for (int i=;i<=n;i++) scanf("%s",mp[i]+);
     int sx,sy,tx,ty,s,t,tot=;
     for (int i=;i<=n;i++)
         for (int j=;j<=m;j++) if (mp[i][j]!='.')
         {
             id[i][j]=++tot;
             if (mp[i][j]=='S') sx=i,sy=j,s=tot;
             if (mp[i][j]=='T') tx=i,ty=j,t=tot;
         }

     if (sx==tx || sy==ty) {puts("-1"); return ;}

     g.clear(tot*+n+m);
     for (int i=;i<=tot;i++) if (i!=s && i!=t) g.insert(i,i+tot,);
     for (int i=;i<=n;i++)
         for (int j=;j<=m;j++) if (id[i][j])
         {
             if (id[i][j]==s)
             {
                 g.insert(s,tot*+i,0x3f3f3f3f);
                 g.insert(tot*+i,s,0x3f3f3f3f);
                 g.insert(s,tot*+n+j,0x3f3f3f3f);
                 g.insert(tot*+n+j,s,0x3f3f3f3f);
             }
             else
             {
                 g.insert(id[i][j]+tot,tot*+i,0x3f3f3f3f);
                 g.insert(tot*+i,id[i][j],0x3f3f3f3f);
                 g.insert(id[i][j]+tot,tot*+n+j,0x3f3f3f3f);
                 g.insert(tot*+n+j,id[i][j],0x3f3f3f3f);
             }
         }
     printf("%d\n",g.Dinic(s,t));
     return ;
 }