poj3621(0-1分数规划问题）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
 
const int maxn=10000+10;
const double eps=0.0001;
const int nil=0x3f3f3f3f;
 
struct my{
       int next,v;
       double w;
};
 
int adj[maxn],adj2[maxn],cnt[maxn],fa,fa2,n,m;
double w[maxn],d[maxn];
queue<int>Q;
bool inq[maxn];
my bian[maxn],bian2[maxn];
 
void myinsert(int u,int v,double w){
     bian[++fa].v=v;
     bian[fa].next=adj[u];
     bian[fa].w=w;
     adj[u]=fa;
}
 
void myinsert2(int u,int v,double w){
     bian2[++fa2].v=v;
     bian2[fa2].next=adj2[u];
     bian2[fa2].w=w;
     adj2[u]=fa2;
}
 
bool spfa(int s){
     for (int i=1;i<=n;i++) d[i]=nil,inq[i]=0,cnt[i]=0;
     while(!Q.empty()) Q.pop();
     Q.push(s);
     d[s]=0;
     inq[s]=true;
     while(!Q.empty()){
        int u=Q.front();Q.pop();
        inq[u]=false;
        for (int i=adj[u];i;i=bian[i].next){
            int v=bian[i].v;
            if(d[u]<nil&&d[v]>d[u]+bian[i].w){
                d[v]=d[u]+bian[i].w;
                if(!inq[v]){
                    inq[v]=true;
                    Q.push(v);
                    if(++cnt[v]>n) return false;
                }
            }
        }
     }
     return true;
}
 
bool check(double mid){
     for (int i=1;i<=n;i++){
        for (int j=adj[i];j;j=bian[j].next){
            bian[j].w=bian2[j].w*mid-w[bian[j].v];
        }
     }
     if(!spfa(1)) return true;
    return false;
}
 
int main(){
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++){
        scanf("%lf",&w[i]);
    }
    int u,v;
    double w;
    for (int i=1;i<=m;i++){
        scanf("%d%d%lf",&u,&v,&w);
        myinsert(u,v,w);
        myinsert2(u,v,w);
    }
    double l=0,r=1000.0,mid;
    while(r-l>eps){
        mid=(l+r)/2.0;
        if(check(mid)) l=mid;
        else r=mid;
        //printf("%.2lf",mid);
    }
    printf("%.2lf
",mid);
return 0;
}

就不用讲了吧

0-1分数规划+判断图的负环