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");elseprintf("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");elseprintf("NO\n");}return 0;
}