P1231 教辅的组成

Posted 2021-03-09 zzctommy

P1231 教辅的组成

嗯，和这题思路一样，我的题解

就是数据范围大了一些。另外注意把边数和点数卡死。我代码的数组就体现了这一过程。

#include<bits/stdc++.h>
using namespace std;
const int inf=1000000007;
const int N=40010;
const int M=20010;
int maxflow,s,t;
int head[N],num_edge=1,cur[N],dep[N];
bool inq[N];
vector<int>g[N];
struct edge {
    int nxt,val,to;
} e[(N+N+N+M+M)<<1];
void add(int from,int to,int val) {
    ++num_edge;
    e[num_edge].nxt=head[from];
    e[num_edge].to=to;
    e[num_edge].val=val;
    head[from]=num_edge;
}
bool bfs() {
    for(int i=1; i<=t; ++i)
        inq[i]=false,cur[i]=head[i],dep[i]=inf;
    queue<int>q;
    q.push(s);
    dep[s]=0;
    inq[s]=true;
    while(!q.empty()) {
        int u=q.front();
        q.pop();
        inq[u]=false;
        for(int i=head[u]; i; i=e[i].nxt) {
            int v=e[i].to;
            if(dep[v]>dep[u]+1&&e[i].val) {
                dep[v]=dep[u]+1;
                if(!inq[v]) {
                    q.push(v);
                    inq[v]=true;
                }
            }
        }
    }
    return dep[t]!=inf;
}
int dfs(int u,int flow) {
    if(u==t) {
        maxflow+=flow;
        return flow;
    }
    int used=0,rlow;
    for(int i=cur[u]; i; i=e[i].nxt) {
        cur[u]=i;
        int v=e[i].to,w=e[i].val;
        if(w&&dep[v]==dep[u]+1) {
            rlow=dfs(v,min(flow-used,w));
            if(rlow) {
                used+=rlow;
                e[i].val-=rlow;
                e[i^1].val+=rlow;
                if(used==flow)break;
            }
        }
    }
    return used;
}
void dinic() {
    while(bfs())dfs(s,inf);
}
int n1,n2,n3,m1,m2;
int main() {
    scanf("%d%d%d",&n1,&n2,&n3);
    s=n2+n1+n3+n1+1;t=s+1; 
    for(int i=n2+1;i<=n2+n1;++i)
        add(i,i+n1+n3,1),add(i+n1+n3,i,0);
    for(int i=1;i<=n2;++i)
        add(s,i,1),add(i,s,0);
    for(int i=n2+n1+1;i<=n2+n1+n3;++i)
        add(i,t,1),add(t,i,0); 
    scanf("%d",&m1);
    for(int i=1,x,y;i<=m1;++i)
    {
        scanf("%d%d",&x,&y);
        add(y,x+n2,1),add(x+n2,y,0);
    }
    scanf("%d",&m2);
    for(int i=1,x,y;i<=m2;++i)
    {
        scanf("%d%d",&x,&y);
        add(x+n2+n1+n3,y+n2+n1,1),add(y+n2+n1,x+n2+n1+n3,0);
    }
    dinic();
    printf("%d
",maxflow);
    return 0;
}