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 '
'
#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
",(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;
}
也可以用dinic
#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;
const int N = 1e6+10, S = N-2, T = N-1, INF = 0x3f3f3f3f;
int n,m;
char s[3000][3000];
struct edge {
int to,w,next;
edge(int to=0,int w=0,int next=0):to(to),w(w),next(next){}
} e[N];
int head[N], dep[N], vis[N], cur[N], cnt=1;
queue<int> Q;
int bfs() {
REP(i,1,n+m) dep[i]=INF,vis[i]=0,cur[i]=head[i];
dep[S]=INF,vis[S]=0,cur[S]=head[S];
dep[T]=INF,vis[T]=0,cur[T]=head[T];
dep[S]=0,Q.push(S);
while (Q.size()) {
int u = Q.front(); Q.pop();
for (int i=head[u]; i; i=e[i].next) {
if (dep[e[i].to]>dep[u]+1&&e[i].w) {
dep[e[i].to]=dep[u]+1;
Q.push(e[i].to);
}
}
}
return dep[T]!=INF;
}
int dfs(int x, int w) {
if (x==T) return w;
int used = 0;
for (int i=cur[x]; i; i=e[i].next) {
cur[x] = i;
if (dep[e[i].to]==dep[x]+1&&e[i].w) {
int f = dfs(e[i].to,min(w-used,e[i].w));
if (f) used+=f,e[i].w-=f,e[i^1].w+=f;
if (used==w) break;
}
}
return used;
}
int dinic() {
int ans = 0;
while (bfs()) ans+=dfs(S,INF);
return ans;
}
void add(int u, int v, int w) {
e[++cnt] = edge(v,w,head[u]);
head[u] = cnt;
e[++cnt] = edge(u,0,head[v]);
head[v] = cnt;
}
int ID(int x,int y) {
return (x-1)*m+y;
}
int main() {
scanf("%d%d",&n,&m);
REP(i,1,n) scanf("%s",s[i]+1);
REP(i,1,n) add(S,i,1);
REP(i,1,m) add(i+n,T,1);
REP(i,1,n) REP(j,1,m) if (s[i][j]=='*') add(i,j+n,INF);
printf("%d
",dinic());
vector<int> ans;
REP(i,1,n) if (dep[i]==INF) ans.pb(i);
printf("%d",(int)ans.size());
for (int x:ans) printf(" %d", x);hr;
ans.clear();
REP(i,1,m) if (dep[i+n]!=INF) ans.pb(i);
printf("%d", (int)ans.size());
for (int x:ans) printf(" %d", x);hr;
}