POJ 3621 Sightseeing Cows | 01分数规划

Posted 2020-10-20 MSPqwq

题目:

http://poj.org/problem?id=3621

---

题解:

二分答案,检查有没有负环

#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 1005
using namespace std;
struct node
{
    int nxt,v;
    double w;
}e[N*5];
int head[N],ecnt,L,P;
double dis[N],fun[N],l,r,mid;
bool vis[N];
void add(int u,int v,int w)
{
    e[++ecnt].v=v;e[ecnt].w=w;e[ecnt].nxt=head[u];head[u]=ecnt;
}
bool spfa_dfs(int u,double val)
{
    vis[u]=1;
    for (int i=head[u],v;i;i=e[i].nxt)
    {
    if (dis[v=e[i].v]>dis[u]+e[i].w*val-fun[u])
    {
        dis[v]=dis[u]+e[i].w*val-fun[u];
        if (!vis[v]) {if (spfa_dfs(v,val)) return 1;}
        else return 1;
     }
    }
    vis[u]=0;
    return 0;
}
bool check(double mid)
{
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    for (int i=1;i<=L;i++)
    if (spfa_dfs(i,mid)) return 1;
    return 0;
}
int main()
{
    scanf("%d%d",&L,&P);
    for (int i=1;i<=L;i++)
    scanf("%lf",&fun[i]),r+=fun[i];
    for (int i=1,u,v,c;i<=P;i++)
    scanf("%d%d%d",&u,&v,&c),add(u,v,c);
    for (int i=1;i<=50;i++)
    {
    mid=(l+r)/2.0;
    if (check(mid)) l=mid;
    else r=mid;
    }
    printf("%.2f\n",l);
    return 0;
}