【BZOJ】1066: [SCOI2007]蜥蜴（最大流）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=1000, M=N*5000, oo=~0u>>1;
int ihead[N], cnt=1, cur[N], gap[N], d[N], p[N], n, m, g, a[22][22];
struct ED { int from, to, cap, next; } e[M];
inline const int id(const int &x, const int &y) { return (x-1)*m+y; }
inline const bool check(const int &i, const int &j, const int &x, const int &y) { return ((i-x)*(i-x)+(j-y)*(j-y))<=g*g; }
inline void add(const int &u, const int &v, const int &w) {
	e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].from=u; e[cnt].cap=w;
	e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].to=u; e[cnt].from=v; e[cnt].cap=0;
}
int isap(const int &s, const int &t, const int &n) {
	for1(i, 0, n) cur[i]=ihead[i];
	int ret=0, i, f, u=s;
	gap[0]=n;
	while(d[s]<n) {
		for(i=cur[u]; i; i=e[i].next) if(e[i].cap && d[u]==d[e[i].to]+1) break;
		if(i) {
			p[e[i].to]=cur[u]=i; u=e[i].to;
			if(u==t) {
				for(f=oo; u!=s; u=e[p[u]].from) f=min(f, e[p[u]].cap);
				for(u=t; u!=s; u=e[p[u]].from) e[p[u]].cap-=f, e[p[u]^1].cap+=f;
				ret+=f;
			}
		}
		else {
			if(! (--gap[d[u]]) ) break;
			d[u]=n; cur[u]=ihead[u];
			for(i=ihead[u]; i; i=e[i].next) if(e[i].cap && d[u]>d[e[i].to]+1) d[u]=d[e[i].to]+1;
			++gap[d[u]];
			if(u!=s) u=e[p[u]].from;
		}
	}
	return ret;
}
int main() {
	read(n); read(m); read(g);
	int s=n*m*2+10, t=s+1, now, tp, ans=0; char c;
	for1(i, 1, n) for1(j, 1, m) {
		for(c=getchar(); c<‘0‘||c>‘9‘; c=getchar());
		tp=c-‘0‘;
		if(tp) {
			now=id(i, j); a[i][j]=tp;
			add(now, now+410, tp);
		}
	}
	for1(i, 1, n) for1(j, 1, m) {
		for(c=getchar(); c!=‘L‘&&c!=‘.‘; c=getchar());
		if(c==‘L‘) add(s, id(i, j), 1), ++ans;
	}
	for1(i, 1, n) for1(j, 1, m) if(a[i][j])
		for(int x=i-g; x<=i+g; ++x) for(int y=j-g; y<=j+g; ++y)
			if(a[x][y] && !(i==x && j==y) && check(i, j, x, y)) add(id(i, j)+410, id(x, y), oo);
	for1(i, 1, g) for1(j, 1, m) add(id(i, j)+410, t, oo), add(id(n-i+1, j)+410, t, oo);
	for1(i, 1, g) for1(j, 1, n) add(id(j, i)+410, t, oo), add(id(j, m-i+1)+410, t, oo);
	print(ans-isap(s, t, t+1));
	return 0;
}