RT，TLE 2个点，不知道咋回事，求大佬帮忙康康

#include<bits/stdc++.h>
using namespace std;
#define id(x,y) ((x)*m+y-m)
#define in(x,y) id(x,y)
#define out(x,y) (id(x,y)+n*m)
const int N=25;
const int V=N*N;
const int E=1e3*V;
const int INF=0x3f3f3f3f;
struct edge{int v,c,nxt;} e[E];
int a[N][N],hd[V],cur[V],lv[V],tot=1,n,m,d,S,T,nV,cnt;
char s[N];
double dist(int x,int y,int X,int Y){return sqrt((X-x)*(X-x)+(Y-y)*(Y-y));}
void add(int u,int v,int c){e[++tot]={v,c,hd[u]};hd[u]=tot;}
bool bfs()
{
	memset(lv,0,sizeof(lv));
	queue<int> q;
	q.push(S);
	lv[S]=1;
	while(!q.empty())
	{
		int u=q.front();q.pop();
		for(int i=hd[u],v,c;i;i=e[i].nxt)
		{
			v=e[i].v; c=e[i].c;
			if(!c||lv[v]) continue;
			q.push(v);
			lv[v]=lv[u]+1;
		}
	}
	return lv[T];
}
int dfs(int u,int f)
{
	if(u==T) return f;
	int sum=0;
	for(int i=cur[u],v,c;i&&f;i=e[i].nxt)
	{
		cur[u]=i;
		v=e[i].v; c=e[i].c;
		if(!c||lv[v]!=lv[u]+1) continue;
		int delta=dfs(v,min(c,f));
		sum+=delta;
		f-=delta;
		e[i].c-=delta;
		e[i^1].c+=delta; 
	}
	return sum;
}
int Dinic()
{
	int maxflow=0;
	while(bfs())
	{
		for(int u=1;u<=nV;++u) cur[u]=hd[u];
		maxflow+=dfs(S,INF);
	}
	return maxflow;
}
int main()
{
	scanf("%d%d%d",&n,&m,&d);
	S=2*n*m+1; T=2*n*m+2; nV=T;
	for(int i=1;i<=n;++i) for(int j=1;j<=m;++j) scanf("%1d",&a[i][j]);
	for(int i=1,c;i<=n;++i) 
		for(int j=1;j<=m;++j)
		{
			c=a[i][j];
			if(c==0) continue; //没有石柱
			add(in(i,j),out(i,j),c);
			add(out(i,j),in(i,j),0); //入->出
			for(int x=1;x<=n;++x)
				for(int y=1;y<=m;++y) if((x!=i||y!=j)&&dist(i,j,x,y)<=d&&a[x][y]) //平面距离<=d且有石柱
					add(out(i,j),in(x,y),INF), add(in(x,y),out(i,j),0);
			if(i<=d||j<=d||n-i+1<=d||m-j+1<=d) 
				add(out(i,j),T,INF),add(T,out(i,j),0); //可以一步跳到界外
		}
	for(int i=1;i<=n;++i)
	{
		scanf("%s",s+1);
		for(int j=1;j<=m;++j) 
			if(s[j]=='L') 
				add(S,in(i,j),1),add(in(i,j),S,0),++cnt;
	}
	printf("%d\n",cnt-Dinic());
	return 0;
}

看上去带当前弧的dinic没有问题。（难道是define过多导致效率低下？）