牛客 216D 消消乐 (二分图最小点覆盖)

大意: 给定棋盘, 每次消除一行或一列, 求最小次数使得消除完所有'*'.

裸的二分图最小点覆盖.

二分图的最小点覆盖等于最大匹配, 输出方案时从所有左部未盖点开始标记交替路上的点, 最后左部所有未标记的点加上右部所有标记的点即为最小点覆盖.

#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 = 1e5+10;
int n, m, clk, vis[N], f[N];
char s[N];
vector<int> g[N];
int dfs(int x) {
	vis[x] = clk;
	for (int y:g[x]) if (vis[y]!=clk) {
		vis[y] = clk;
		if (!f[y]||dfs(f[y])) return f[y]=x;
	}
	return 0;
}
int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,n) { 
		scanf("%s",s+1);
		REP(j,1,m) if (s[j]=='*') { 
			g[i].pb(j+n);
			g[j+n].pb(i);
		}
	}
	REP(i,1,n) ++clk, dfs(i);
	REP(i,n+1,n+m) f[f[i]]=i;
	++clk;
	REP(i,1,n) if (!f[i]) dfs(i);
	vector<int> raw, col;
	REP(i,1,n) if (vis[i]!=clk) raw.pb(i);
	REP(i,n+1,n+m) if (vis[i]==clk) col.pb(i-n);
	printf("%d\n",(int)raw.size()+(int)col.size());
	printf("%d",(int)raw.size());
	for (int i:raw) printf(" %d",i);hr;
	printf("%d",(int)col.size());
	for (int i:col) printf(" %d",i);hr;
}