gift 分数规划的最大权闭合子图

题目大意：

N个物品,物品间有M组关系,每个物品有一个ai的代价,满足关系后会得到bi的值

求 max(sigma(bi)/sigma(ai))

题解：

很明显的最大权闭合子图，只不过需要处理分数.

这里二分一个答案g

然后直接求sigma(b[i])-sigma(a[i]*g)即可

其中把图中的ai都改成ai*g即可

注意整数处理

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
using namespace std;
const int N=4005,M=200005,INF=2e8;
int n,m;double SUM=0,ep=0.0000001;
int gi(){
    int str=0;char ch=getchar();
    while(ch>'9' || ch<'0')ch=getchar();
    while(ch>='0' && ch<='9')str=(str<<1)+(str<<3)+ch-48,ch=getchar();
    return str;
}
int val[N];
struct node{
    int x,y,v;
}c[M];
int head[(N+M)],num=1;
struct Lin{
    int next,to;double dis;
}a[(N+M)<<2];
void init(int x,int y,double dis){
    a[++num].next=head[x];
    a[num].to=y;a[num].dis=dis;
    head[x]=num;
    a[++num].next=head[y];
    a[num].to=x;a[num].dis=0;
    head[y]=num;
}
int S=0,T,q[N+M],dep[N+M];
bool bfs(){
    for(int i=0;i<=T;i++)dep[i]=0;
    q[1]=S;dep[S]=1;
    int t=0,sum=1,u,x;
    while(t!=sum){
        t++;x=q[t];
        for(RG int i=head[x];i;i=a[i].next){
            u=a[i].to;
            if(dep[u] || a[i].dis<ep)continue;
            dep[u]=dep[x]+1;
            q[++sum]=u;
        }
    }
    return dep[T];
}
double dinic(int x,double tot){
    if(x==T || !tot)return tot;
    int u;double tmp,sum=0;
    for(RG int i=head[x];i;i=a[i].next){
        u=a[i].to;
        if(a[i].dis<ep || dep[u]!=dep[x]+1)continue;
        tmp=dinic(u,min(tot,a[i].dis));
        a[i].dis-=tmp;a[i^1].dis+=tmp;
        sum+=tmp;tot-=tmp;
        if(!tot)break;
    }
    if(sum<ep)dep[x]=0;
    return sum;
}
double maxflow(){
    double tot=0,tmp;
    while(bfs()){
        tmp=dinic(S,INF);
        while(tmp>=ep)tot+=tmp,tmp=dinic(S,INF);
    }
    return tot;
}
void Clear(){
    num=1;
    memset(head,0,sizeof(head));
}
bool check(double g){
    Clear();
    T=n+m+1;
    for(RG int i=1;i<=n;i++)init(S,i,g*(double)val[i]);
    for(RG int i=1;i<=m;i++){
        init(c[i].x,i+n,INF);
        init(c[i].y,i+n,INF);
        init(i+n,T,c[i].v);
    }
    double pap=maxflow();
    pap=SUM-pap;
    if(pap>=ep)return true;
    return false;
}
void work(){
    n=gi();m=gi();
    for(RG int i=1;i<=n;i++)val[i]=gi();
    for(RG int i=1;i<=m;i++)c[i].x=gi(),c[i].y=gi(),c[i].v=gi(),SUM+=c[i].v;
    double l=1,r=SUM,mid,ans;
    while(l<=r){
        mid=(l+r)/2;
        if(check(mid)){
            ans=mid;
            l=mid+ep;
        }
        else r=mid-ep;
    }
    printf("%d
",(int)ans);
}
int main()
{
    freopen("gift.in","r",stdin);
    freopen("gift.out","w",stdout);
    work();
    return 0;
}