AC日记——[USACO5.4]奶牛的电信Telecowmunication 洛谷 P1345

Posted 2020-09-29 Only U - IU

[USACO5.4]奶牛的电信Telecowmunication

 

思路：

　　水题；

 

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 205
#define maxm 20005
#define INF 0x3f3f3f3f
int head[maxn],cnt=1,n,m,E[maxm],V[maxm],F[maxm],s,t;
int que[maxm<<2],deep[maxn],c1,c2;
inline void in(int &now)
{
    char Cget=getchar();now=0;
    while(Cget>‘9‘||Cget<‘0‘) Cget=getchar();
    while(Cget>=‘0‘&&Cget<=‘9‘)
    {
        now=now*10+Cget-‘0‘;
        Cget=getchar();
    }
}
inline void edge_add(int u,int v,int f)
{
    E[++cnt]=head[u],V[cnt]=v,F[cnt]=f,head[u]=cnt;
    E[++cnt]=head[v],V[cnt]=u,F[cnt]=0,head[v]=cnt;
}
bool bfs()
{
    memset(deep,-1,sizeof(deep));
    int h=0,tail=1,now;que[0]=s,deep[s]=0;
    while(h<tail)
    {
        now=que[h++];
        for(int i=head[now];i;i=E[i])
        {
            if(deep[V[i]]<0&&F[i])
            {
                deep[V[i]]=deep[now]+1;
                if(V[i]==t) return true;
                que[tail++]=V[i];
            }
        }
    }
    return false;
}
int flowing(int now,int flow)
{
    if(t==now||flow<=0) return flow;
    int oldflow=0,pos;
    for(int i=head[now];i;i=E[i])
    {
        if(!F[i]||deep[V[i]]!=deep[now]+1) continue;
        pos=flowing(V[i],min(flow,F[i]));
        flow-=pos,oldflow+=pos,F[i]-=pos,F[i^1]+=pos;
        if(!flow) return oldflow;
    }
    if(!oldflow) deep[now]=-1;
    return oldflow;
}
int main()
{
    in(n),in(m),in(c1),in(c2),s=c1+n,t=c2;
    for(int i=1;i<=n;i++) edge_add(i,i+n,1);
    while(m--)
    {
        in(c1),in(c2);
        edge_add(c1+n,c2,INF);
        edge_add(c2+n,c1,INF);
    }
    int ans=0;
    while(bfs()) ans+=flowing(s,INF);
    cout<<ans;
    return 0;
}