洛谷P3386 【模板】二分图匹配
板子,直接用dinic
1 //minamoto 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<queue> 6 #define inf 0x3f3f3f3f 7 using namespace std; 8 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) 9 char buf[1<<21],*p1=buf,*p2=buf; 10 inline int read(){ 11 #define num ch-'0' 12 char ch;bool flag=0;int res; 13 while(!isdigit(ch=getc())) 14 (ch=='-')&&(flag=true); 15 for(res=num;isdigit(ch=getc());res=res*10+num); 16 (flag)&&(res=-res); 17 #undef num 18 return res; 19 } 20 const int N=2005,M=2100005; 21 int ver[M],Next[M],head[N],edge[M],tot=1; 22 int dep[N],cur[N],s,t; 23 int n,m,k; 24 queue<int> q; 25 inline void add(int u,int v,int e){ 26 ver[++tot]=v,Next[tot]=head[u],head[u]=tot,edge[tot]=e; 27 ver[++tot]=u,Next[tot]=head[v],head[v]=tot,edge[tot]=0; 28 } 29 bool bfs(){ 30 memset(dep,-1,sizeof(dep)); 31 while(!q.empty()) q.pop(); 32 for(int i=s;i<=t;++i) cur[i]=head[i]; 33 q.push(s),dep[s]=0; 34 while(!q.empty()){ 35 int u=q.front();q.pop(); 36 for(int i=head[u];i;i=Next[i]){ 37 int v=ver[i]; 38 if(dep[v]<0&&edge[i]){ 39 dep[v]=dep[u]+1,q.push(v); 40 if(v==t) return true; 41 } 42 } 43 } 44 return false; 45 } 46 int dfs(int u,int limit){ 47 if(!limit||u==t) return limit; 48 int flow=0,f; 49 for(int i=cur[u];i;i=Next[i]){ 50 int v=ver[i];cur[u]=i; 51 if(dep[v]==dep[u]+1&&(f=dfs(v,min(limit,edge[i])))){ 52 flow+=f,limit-=f; 53 edge[i]-=f,edge[i^1]+=f; 54 if(!limit) break; 55 } 56 } 57 return flow; 58 } 59 int dinic(){ 60 int flow=0; 61 while(bfs()) flow+=dfs(s,inf); 62 return flow; 63 } 64 int main(){ 65 n=read(),m=read(),k=read(); 66 s=0,t=n+m+1; 67 for(int i=1;i<=n;++i) add(s,i,1); 68 for(int i=n+1;i<=n+m;++i) add(i,t,1); 69 for(int i=1;i<=k;++i){ 70 int u=read(),v=read(); 71 if(v>m) continue; 72 add(u,v+n,1); 73 } 74 printf("%d\n",dinic()); 75 return 0; 76 }
深深地明白自己的弱小