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; }
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 '\n' #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\n",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; }