POJ 1860 Currency Exchange

英语太渣读了半天，理解了题意就好办了……Bellman_ford算法。在n-1次松弛后，如果依然存在满足松弛的情况返回1.说简单点就是判读是否存在正环。

#include<stdio.h>
#include<string.h>
#include<cstring>
#include<string>
#include<math.h>
#include<queue>
#include<algorithm>
#include<iostream>
#include<stdlib.h>
#include<cmath>

#define INF 0x3f3f3f3f
#define MAX 1000005

using namespace std;

struct node
{
    int a,b;
    double e,c;
}Map[MAX];

int n,m,s,k;

double dist[MAX],v;

int bellman_ford()
{
    int i,j,ok;
    memset(dist,0,sizeof(dist));

    dist[s]=v;

    for(i=1;i<n;i++)
    {
        ok=0;
        for(j=1;j<k;j++)
        {
            if(dist[Map[j].b] < (dist[Map[j].a] - Map[j].c)*Map[j].e)
            {
                dist[Map[j].b] = (dist[Map[j].a] - Map[j].c)*Map[j].e;
                ok=1;
            }
        }

        if(!ok)
            break;
    }

    for(j=1;j<k;j++)
    {
        if(dist[Map[j].b] < (dist[Map[j].a] - Map[j].c)*Map[j].e)
            return 1;
    }

    return 0;
}

int main()
{
    int i,j,a,b;

    double e1,c1,e2,c2;

    while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF)
    {
        k=1;

        for(i=1;i<=m;i++)
        {
            scanf("%d%d%lf%lf%lf%lf",&a,&b,&e1,&c1,&e2,&c2);

            Map[k].a=a;
            Map[k].b=b;
            Map[k].e=e1;
            Map[k++].c=c1;
            Map[k].a=b;
            Map[k].b=a;
            Map[k].e=e2;
            Map[k++].c=c2;
        }

        int ok=bellman_ford();

        if(ok)
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}