AtCoder Regular Contest 074 F - Lotus Leaves

题目传送门：https://arc074.contest.atcoder.jp/tasks/arc074_d

题目大意：

给定一个\(H×W\)的网格图，o是可以踩踏的点，.是不可踩踏的点。

现有一人在S处，向T移动，若此人现在在\((i,j)\)上，那么下一步他可以移动到​\((i,k)\)或\((k,j)\)上，\(k\)任意

问最少需要将多少个o改成.，可以使这个人无法从S到达T，输出最少需要更改的数目；如果无论如何都不能使这个人无法从S到T，则输出\(-1\)

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这个模型就是最小割啊……我们对于每个点拆点，然后对于所有o，入点向出点连一条流量为1的边，S和T的话，入点向出点连一条流量为\(\infty\)的边；对于移动，我们由某个点的出点向其能到达的入点连一条流量为\(\infty\)的边，直接最小割即可

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e2,M=6e6;
char map[N+10][N+10];
int pre[M+10],now[N*N*2+10],child[M+10],val[M+10];
int h[N*N*2+10],deep[N*N*2+10];
bool vis[N*N*2+10];
int n,m,S,T,tot=1;
int Get(int x,int y){return (x-1)*m+y;}
void join(int x,int y,int z){pre[++tot]=now[x],now[x]=tot,child[tot]=y,val[tot]=z;}
void insert(int x,int y,int z){join(x,y,z),join(y,x,0);}
bool bfs(int x){
	int head=1,tail=1;
	memset(deep,255,sizeof(deep));
	deep[h[1]=x]=0;
	for (;head<=tail;head++){
		int Now=h[head];
		for (int p=now[Now],son=child[p];p;p=pre[p],son=child[p]){
			if (!~deep[son]&&val[p]){
				deep[h[++tail]=son]=deep[Now]+1;
				if (son==T)	return 1;
			}
		}
	}
	return 0;
}
int dfs(int x,int v){
	if (x==T)	return v;
	int res=0;
	for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){
		if (deep[son]>deep[x]&&val[p]){
			int k=dfs(son,min(v,val[p]));
			val[p]-=k,v-=k;
			val[p^1]+=k,res+=k;
			if (!v)	break;
		}
	}
	if (!res)	deep[x]=-1;
	return res;
}
int main(){
	n=read(),m=read(),S=2*n*m+1,T=S+1;
	for (int i=1;i<=n;i++){
		scanf("%s",map[i]+1);
		for (int j=1;j<=m;j++){
			if (map[i][j]=='.')	continue;
			if (map[i][j]!='o'){
				insert(Get(i,j),Get(i,j)+n*m,inf);
				if (map[i][j]=='S')	insert(S,Get(i,j),inf);
				if (map[i][j]=='T')	insert(Get(i,j)+n*m,T,inf);
				map[i][j]='o';
			}else	insert(Get(i,j),Get(i,j)+n*m,1);
		}
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			if (map[i][j]=='o'){
				for (int k=1;k<=n;k++)	if (map[k][j]=='o'&&k!=i)	insert(Get(i,j)+n*m,Get(k,j),inf);
				for (int k=1;k<=m;k++)	if (map[i][k]=='o'&&k!=j)	insert(Get(i,j)+n*m,Get(i,k),inf);
			}
		}
	}
	int Ans=0;
	while (bfs(S))	Ans+=dfs(S,inf);
	printf(Ans>=inf?"-1\n":"%d\n",Ans);
	return 0;
}