落谷 P1231 教辅的组成

题目大意：

恩呃呃呃…给你A,B,C三类的物品，给出A,B；B,C的匹配关系，一个A,B,C类物品在一起就算一套物品。求出最多可以有多少套物品。

思路：

最大流的问题，建立一个超级源点和超级汇点，然后A-B-B’-C连边，从源点跑一次网络流然后就可以求出答案了

程序：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<queue>#define MAXN 1000010#define MAX 999999999using namespace std;int n,m,k,s,t,c=2,ans=0;int head[MAXN],next[MAXN],go[MAXN],w[MAXN],deep[MAXN];inline int read(){    int date=0,w=1; char c=0;    while (c<'0'||c>'9'){if (c=='-') w=-1;c=getchar();}    while (c>='0'&&c<='9'){date=date*10+c-'0';c=getchar();}    return date*w;}void add(int x,int y,int p){        go[c]=y; next[c]=head[x];    w[c]=p;    head[x]=c++;    go[c]=x; next[c]=head[y];    w[c]=0;    head[y]=c++;}bool bfs(){    int u,v;    queue<int>q;    memset(deep,0,sizeof(deep));    deep[s]=1;    q.push(s);    while (!q.empty()){        u=q.front();        q.pop();        for (int i=head[u];i;i=next[i]){            v=go[i];            if (w[i]&&!deep[v]){                deep[v]=deep[u]+1;                if (v==t) return true;                q.push(v);            }        }    }    return false;}int dfs(int x,int mi){    if (x==t) return mi;    int v,sum,cost=0;    for (int i=head[x];i;i=next[i]){        v=go[i];        if (w[i]&&deep[v]==deep[x]+1){            sum=dfs(v,min(mi-cost,w[i]));        if (sum>0){            w[i]-=sum;            w[i^1]+=sum;            cost+=sum;            if (cost==mi) break;        } else deep[v]=-1;        }    }    return cost;}int dinic(){    int ans=0;    while (bfs()) ans+=dfs(s,MAX);    return ans;}int main(){    int f,u,v;    n=read(); m=read(); k=read();    f=read();    for (int i=1;i<=f;i++){        u=read(); v=read();        add(v,u+m,1);    }    f=read();    for (int i=1;i<=f;i++){        u=read(); v=read();        add(u+n+m,v+n+n+m,1);    }    for (int i=1;i<=n;i++) add(i+m,i+m+n,1);    for (int i=1;i<=m;i++) add(n+m+k+n+1,i,1);    for (int i=1;i<=k;i++) add(i+n+n+m,n+n+m+k+2,1);    s=n+m+k+n+1; t=n+n+m+k+2;    printf("%d\n",dinic());    return 0;}