# ARC 074 F - Lotus Leaves ( Flow, Mincut )

F: Lotus Leaves - AtCoder Regular Contest 074 | AtCoder

H * W 的棋盤。

2≤H,W≤100
aij は ., o, S, T のどれかである。
aij のうち S はちょうど 1 個存在する。
aij のうち T はちょうど 1 個存在する。

```#include <bits/stdc++.h>
using namespace std;

template< class T >
struct Dinic {
static const int MAXV = 10000;
static const T INF = 0x3f3f3f3f;
struct Edge {
int v;
T f;
int re;
Edge( int _v, T _f, int _re ): v( _v ), f( _f ), re( _re ) {}
};
int n, s, t, level[ MAXV ];
vector< Edge > E[ MAXV ];
int now[ MAXV ];
Dinic( int _n, int _s, int _t ): n( _n ), s( _s ), t( _t ) {}
void add_edge( int u, int v, T f, bool bidirectional = false ) {
E[ u ].emplace_back( v, f, E[ v ].size() );
E[ v ].emplace_back( u, 0, E[ u ].size() - 1 );
if( bidirectional ) {
E[ v ].emplace_back( u, f, E[ u ].size() - 1 );
}
}
bool BFS() {
memset( level, -1, sizeof( level ) );
queue< int > que;
que.emplace( s );
level[ s ] = 0;
while( not que.empty() ) {
int u = que.front();
que.pop();
for( auto it: E[ u ] ) {
if( it.f > 0 and level[ it.v ] == -1 ) {
level[ it.v ] = level[ u ] + 1;
que.emplace( it.v );
}
}
}
return level[ t ] != -1;
}
T DFS( int u, T nf ) {
if( u == t ) return nf;
T res = 0;
while( now[ u ] < E[ u ].size() ) {
Edge &it = E[ u ][ now[ u ] ];
if( it.f > 0 and level[ it.v ] == level[ u ] + 1 ) {
T tf = DFS( it.v, min( nf, it.f ) );
res += tf; nf -= tf; it.f -= tf;
E[ it.v ][ it.re ].f += tf;
if( nf == 0 ) return res;
} else ++now[ u ];
}
if( not res ) level[ u ] = -1;
return res;
}
T flow( T res = 0 ) {
while( BFS() ) {
T temp;
memset( now, 0, sizeof( now ) );
while( temp = DFS( s, INF ) ) {
res += temp;
res = min( res, INF );
}
}
return res;
}
};

const int MAXH = 100;
const int MAXW = 100;

int H, W;
string G[ MAXH ];

signed main() {
ios::sync_with_stdio( 0 );
cin >> H >> W;
for( int i = 0; i < H; ++i ) {
cin >> G[ i ];
}
int SOURCE = H + W;
int SINK = H + W + 1;
Dinic< int > din( H + W + 2, SOURCE, SINK );
for( int i = 0; i < H; ++i ) {
for( int j = 0; j < W; ++j ) {
if( G[ i ][ j ] == 'S' ) {
din.add_edge( SOURCE, H + j, din.INF );
} else if( G[ i ][ j ] == 'T' ) {