LG1345 「USACO5.4」Telecowmunication 最小割

问题描述

LG1345

题解

点边转化，最小割，完事。

\(\mathrm{Code}\)

#include<bits/stdc++.h>
using namespace std;

template <typename Tp>
void read(Tp &x){
    x=0;char ch=1;int fh;
    while(ch!='-'&&(ch>'9'||ch<'0')) ch=getchar();
    if(ch=='-') ch=getchar(),fh=-1;
    else fh=1;
    while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
    x*=fh;
}

const int maxn=1007;
const int maxm=10007;
const int INF=0x3f3f3f3f;

int n,m,xx,yy;
int Head[maxn],to[maxm],w[maxm],Next[maxm],tot=1;

void add(int x,int y,int z){
    to[++tot]=y,Next[tot]=Head[x],Head[x]=tot,w[tot]=z;//错误笔记：如果我加边再不计边权，我就*****
}

int S,T,d[maxn];

bool bfs(){
    memset(d,0,sizeof(d));
    queue<int>q;q.push(S);d[S]=1;
    while(!q.empty()){
        int x=q.front();q.pop();
        for(int i=Head[x];i;i=Next[i]){
            int y=to[i];
            if(d[y]||!w[i]) continue;
            q.push(y);d[y]=d[x]+1;
            if(y==T) return true;
        }
    }
    return false;
}

int dfs(int x,int flow){
    if(x==T) return flow;
    int rest=flow;
    for(int i=Head[x];i&&rest;i=Next[i]){
        int y=to[i];
        if(d[y]!=d[x]+1||!w[i]) continue;
        int k=dfs(y,min(rest,w[i]));
        if(!k) d[y]=0;
        else{
            w[i]-=k,w[i xor 1]+=k;
            rest-=k;
        }
    }
    return flow-rest;
}

int main(){
    read(n);read(m);read(xx);read(yy);
    for(int i=1,x,y;i<=m;i++){
        read(x);read(y);
        add(x+n,y,INF);add(y,x+n,0);
        add(y+n,x,INF);add(x,y+n,0);
    }
    for(int i=1;i<=n;i++){
        add(i,i+n,1);add(i+n,i,0);
    }
    S=xx+n,T=yy;
    int ans=0;
    while(bfs()){
        int t;
        while(t=dfs(S,0x3f3f3f3f)) ans+=t;
    }
    printf("%d\n",ans);
    return 0;
}

标签：ch,return,int,USACO5.4,flow,rest,read,Telecowmunication,LG1345
来源： https://www.cnblogs.com/liubainian/p/11779809.html