poj_1860

此题是求“最长路&&回路”，最长路的话我们只要把最短路的条件改一改就行，既然要求回路，那么我们就用bellman_ford来求，因为每求一个点后，都有可能使得最长路松弛，所以我们遍历n-1遍所有的边，如果之后还有可以松弛的边，那么说明这个图中存在着回路，也就是对于本题，货币可以无限的换，无限的增长。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>//#define ONLINE_JUDGE#define eps 1e-10#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll __int64const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;int n,m,k;int s;double v;#define N 8080struct Edge{    int u,v;    double rate,com;    //记录一条边的两个点，以及所代表的汇率和手续费}e[550];bool belman_Ford(){    double dis[550];            //    memset(dis,0,sizeof dis);   //此题要求的是能够换的钱越多越好，所以我们初始化全为0    dis[s] = v; //dis[s]=v    for(int i=1;i<n;i++){        for(int j=0;j<k;j++){            if(dis[e[j].v]<(dis[e[j].u]-e[j].com)*e[j].rate)//最长路，所以松弛的条件改变                dis[e[j].v] = (dis[e[j].u]-e[j].com)*e[j].rate;        }    }    for(int i=0;i<k;i++)    //如果还有回路，说明货币数量可以无限增大        if(dis[e[i].v]<(dis[e[i].u]-e[i].com)*e[i].rate)            return true;    return false;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif    while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF){        int u,v;        double rate1,com1,rate2,com2;        k=0;        for(int i=0;i<m;i++){            scanf("%d%d%lf%lf%lf%lf",&u,&v,&rate1,&com1,&rate2,&com2);            e[k].u = u;            e[k].v = v;            e[k].rate = rate1;            e[k++].com = com1;            e[k].u = v;            e[k].v = u;            e[k].rate = rate2;            e[k++].com = com2;        }        if(belman_Ford()){            printf("YES\n");        }else            printf("NO\n");    }return 0;}