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题目链接:http://poj.org/problem?id=1860
题目大意:给你一些兑换方式,问你能否通过换钱来赚钱?
使用ford算法,当出现赚钱的时候就返回YES,如果不能赚钱,则返回NO
应该是可以停下来的,但是我不会分析复杂度,谁来教教我?
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <vector> 5 #include <map> 6 #include <set> 7 #include <bitset> 8 #include <cmath> 9 #include <numeric> 10 #include <iterator> 11 #include <iostream> 12 #include <cstdlib> 13 using namespace std; 14 #define PB push_back 15 #define MP make_pair 16 #define SZ size() 17 #define ST begin() 18 #define ED end() 19 #define CLR clear() 20 #define ZERO(x) memset((x),0,sizeof(x)) 21 typedef long long LL; 22 typedef unsigned long long ULL; 23 typedef pair<int,int> PII; 24 const double EPS = 1e-8; 25 26 struct EDGE{ 27 int from,to; 28 double r,c; 29 EDGE(int _from = 0, int _to =0 , double _r = 0, double _c = 0 ) : 30 from(_from),to(_to),r(_r),c(_c) {} 31 }; 32 33 vector<EDGE> vec; 34 int N,M,S; 35 double V; 36 double d[111]; 37 38 bool floyd(int s){ 39 d[s] = V; 40 bool flag = false; 41 while(true){ 42 flag = false; 43 for(int i=0;i<vec.SZ;i++){ 44 const EDGE& e = vec[i]; 45 if(d[e.from]>0.0 && d[e.to]<(d[e.from]-e.c)*e.r){ 46 d[e.to]=(d[e.from]-e.c)*e.r; 47 flag = true; 48 } 49 } 50 if(d[s]>V) return true; 51 if(!flag) { 52 break; 53 } 54 } 55 return d[s]>V; 56 } 57 58 59 int main(){ 60 while(EOF!=scanf("%d%d%d%lf",&N,&M,&S,&V)){ 61 vec.CLR; 62 ZERO(d); 63 int from,to; 64 double rab,cab,rba,cba; 65 for(int i=0;i<M;i++){ 66 scanf("%d%d%lf%lf%lf%lf",&from,&to,&rab,&cab,&rba,&cba); 67 vec.PB(EDGE(from,to,rab,cab)); 68 vec.PB(EDGE(to,from,rba,cba)); 69 } 70 puts(floyd(S)?"YES":"NO"); 71 } 72 return 0; 73 }
[poj1860] Currency Exchange (bellman-ford算法)
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原文地址:http://www.cnblogs.com/llkpersonal/p/4393246.html