链接:poj 1860

题意：给定n中货币。以及它们之间的税率。A货币转化为B货币的公式为 B=(V-Cab)*Rab,当中V为A的货币量，

求货币S通过若干此转换，再转换为原本的货币时是否会添加

分析：这个题就是推断是否存在正权回路。能够用bellman-ford算法，只是松弛条件相反

也能够用SPFA算法，推断经过转换后，转换为原本货币的值是否比原值大、、、


bellman-ford    0MS

#include<stdio.h>
#include<string.h>
struct stu
{
    int a,b;
    double r,c;
}edge[205];
double v,dis[105];
int s;
int bellmanford(int n,int m)
{
    int i,j,flag=0;
    memset(dis,0,sizeof(dis));
    dis[s]=v;
    for(i=1;i<=n-1;i++)
        for(j=1;j<=m;j++)
            if(dis[edge[j].a]&&(dis[edge[j].a]-edge[j].c)*edge[j].r>dis[edge[j].b])
                dis[edge[j].b]=(dis[edge[j].a]-edge[j].c)*edge[j].r;
    for(j=1;j<=m;j++)
        if(dis[edge[j].a]&&(dis[edge[j].a]-edge[j].c)*edge[j].r>dis[edge[j].b]){
            flag=1;
            break;
        }
    return flag;
}
int main()
{
    int i,j,l,r,n,m,flag;
    while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF){
        j=1;
        for(i=1;i<=m;i++){
            scanf("%d%d",&l,&r);
            scanf("%lf%lf",&edge[j].r,&edge[j].c);
            edge[j].a=l;
            edge[j].b=r;
            j++;
            edge[j].a=r;
            edge[j].b=l;
            scanf("%lf%lf",&edge[j].r,&edge[j].c);
            j++;
        }
        flag=bellmanford(n,2*m);
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

SPFA+邻接表  16MS

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
struct stu
{
    int a,b;
    double r,c;
}edge[205];
double v,dis[105];
int s,first[205],next[205],vis[105];
int SPFA(int n,int m)
{
    int i,pos;
    queue<int> q;
    memset(dis,0,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dis[s]=v;
    q.push(s);
    vis[s]=1;
    while(!q.empty()){
        pos=q.front();
        q.pop();
        vis[pos]=0;
        i=first[pos];
        while(i!=-1){
            if((dis[pos]-edge[i].c)*edge[i].r>dis[edge[i].b]){
                dis[edge[i].b]=(dis[pos]-edge[i].c)*edge[i].r;
                if(!vis[edge[i].b]){
                    q.push(edge[i].b);
                    vis[edge[i].b]=1;
                }
            }
            i=next[i];
        }
        if(dis[s]>v)
            return 1;
    }
    return 0;
}
int main()
{
    int i,j,l,r,n,m,flag;
    while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF){
        j=1;
        for(i=1;i<=m;i++){
            scanf("%d%d",&l,&r);
            scanf("%lf%lf",&edge[j].r,&edge[j].c);
            edge[j].a=l;
            edge[j].b=r;
            j++;
            edge[j].a=r;
            edge[j].b=l;
            scanf("%lf%lf",&edge[j].r,&edge[j].c);
            j++;
        }
        memset(first,-1,sizeof(first));
        for(i=1;i<=2*m;i++){
            next[i]=first[edge[i].a];
            first[edge[i].a]=i;
        }
        flag=SPFA(n,2*m);
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}