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
#includeSPFA+邻接表 16MS#include 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;}
#include#include #include 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 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;}