poj1860

经典bellman-ford,目标是求解最长路，(a-c)*r前加负号则转化为求解最短路，更简单的是用求解最短路的方式求解最长路（这举可行是题目原因）

#include <stdio.h>
#include <string.h>
#define MAXN 300
#define INF -(1<<29)
int n,m,s;
double v;
struct node
{
    double r,c;
}map[MAXN][MAXN];
double d[MAXN];
void bellman()
{
    for(int i=1;i<=n;d[i++]=INF);
    d[s]=v;
    for(int k=1;k<=n-1;++k)
        for(int i=1;i<=n;++i)
            for(int j=1;j<=n;++j)
                if(d[j]<(d[i]-map[i][j].c)*map[i][j].r)
                    d[j]=(d[i]-map[i][j].c)*map[i][j].r;
}
int main()
{
    while(scanf("%d %d %d %lf",&n,&m,&s,&v)==4)
    {
        for(int i=0,a,b;i<m;++i)
        {
            scanf("%d %d",&a,&b);
            scanf("%lf %lf %lf %lf",&map[a][b].r,&map[a][b].c,&map[b][a].r,&map[b][a].c);
        }
        bellman();
        bool f=true;
        for(int i=1;i<=n;++i)
            for(int j=1;j<=n;++j)
                if(d[j]<(d[i]-map[i][j].c)*map[i][j].r)
                {
                    f=false;
                    goto answer;
                }
        answer:
        printf((f?"NO\n":"YES\n"));
    }
    return 0;
}