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 '\n'

#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\n", ans);

	}

}