poj 2226 Muddy Fields (二分图)

大意:给定n*m网格, 每个格子为泥地或草地, 可以用一些长度任意宽度为1的木板盖住泥地, 要求不能盖到草地, 求最少要多少块木板能盖住所有泥地.

最小点覆盖板子题, 建图跑最大匹配即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 110;
int n, m, clk, f[N*N], vis[N*N];
int col[N*N], raw[N*N], f1[N][N], f2[N][N];
vector<int> g[N*N];
char a[N][N];
int has(int i, int j) {return (i-1)*m+j;}
void add(int i, int j) {
	g[i].pb(j),g[j].pb(i);
}
int dfs(int x) {
	for (vector<int>::iterator it = g[x].begin(); it!=g[x].end(); ++it) {
		int y = *it;
		if (vis[y]!=clk) {
			vis[y] = clk;
			if (!f[y]||dfs(f[y])) return f[y]=x;
		}
	}
	return 0;
}
int main() {
	while (~scanf("%d%d", &n, &m)) {
		REP(i,1,n) scanf("%s", a[i]+1);
		*col = *raw = 0;
		REP(i,1,2*n*m) f[i]=0,g[i].clear();
		REP(i,1,n) {
			REP(j,1,m) if (a[i][j]=='*') {
				col[++*col] = has(i,j);
				while (a[i][j]=='*') f1[i][j++]=col[*col];
				--j;
			}
		}
		REP(j,1,m) {
			REP(i,1,n) if (a[i][j]=='*') {
				raw[++*raw] = has(i,j)+has(n,m);
				while (a[i][j]=='*') f2[i++][j]=raw[*raw];
				--i;
			}
		}
		REP(i,1,n) REP(j,1,m) if (a[i][j]=='*') add(f1[i][j],f2[i][j]);
		int ans = 0;
		REP(i,1,*col) {
			++clk;
			if (dfs(col[i])) ++ans;
		}
		printf("%d
", ans);
	}
}