牛客 216D 消消乐 (二分图最小点覆盖)-白红宇

牛客 216D 消消乐 (二分图最小点覆盖)

阅读量：5237 次

发布时间：2019-06-14

本文共 3961 字，大约阅读时间需要 13 分钟。

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 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           #include 
            
             #include 
             
              #include 
              
               #include 
               
                #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<
                
                 <<' ';hr;})using namespace std;typedef long long ll;typedef pair
                 
                   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;}//headconst int N = 1e5+10;int n, m, clk, vis[N], f[N];char s[N];vector
                  
                    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
                   
                     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 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           #include 
            
             #include 
             
              #include 
              
               #include 
               
                #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<
                
                 <<' ';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
                 
                   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
                  
                    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;}