【POJ】1860 Currency Exchange

真是气skr人。。没把d[]换成double。。。de了一上午的bug//

记得用G++提交啊

题意：告诉你n个点，m条路。起始点s,还有初始金额money。每条路表示从a->b的汇率和佣金以及b->a的汇率和佣金。你在该点所得是（本金-佣金）*汇率。问你这个人能不能赚钱。

题解：spfa套一下//。记得d[]换成double。具体的看看代码。QWQ。

代码：

 1 #include<iostream>
 2 #include<stack>
 3 #include<vector>
 4 #include<queue>
 5 #include<algorithm>
 6 #include<cstdio>
 7 using namespace std;
 8 const int maxn = 105;
 9 
10 
11 struct node{
12     int to;
13     double r;
14     double c;
15 };
16 
17 vector< node > e[maxn];
18 
19 int n,m,num;
20 double money;
21 double d[maxn];
22 int inq[maxn];
23 
24 bool spfa(int s){
25     for(int i = 1; i <= n ;i++){
26         inq[i] = d[i] = 0;
27     }
28     queue<int>Q;
29     Q.push(s);
30     d[s] = money;
31     inq[s] = 1;
32     while( !Q.empty() ){
33         int now = Q.front();
34         Q.pop();
35         inq[now] = 0;
36         for(int i = 0; i < e[now].size() ; i++){
37             double rate = e[now][i].r;
38             double commis = e[now][i].c;
39             int v = e[now][i].to;
40 
41             if(d[v] < (d[now] - commis) * rate){
42                 d[v] = (d[now] - commis) * rate;
43                 if(inq[v] == 0){
44                     inq[v] = 1;
45                     Q.push(v);
46                 }   
47             }
48 
49             if(d[s] > money){
50                 return true;
51             }
52         }
53 
54 
55     }
56     return false;
57 }
58 
59 int main() {
60     scanf("%d%d%d%lf",&n,&m,&num,&money);
61     int x,y;
62     double r,c;
63     while(m--){
64         scanf("%d%d%lf%lf",&x,&y,&r,&c);
65         e[x].push_back((node){y,r,c});
66         scanf("%lf%lf",&r,&c);
67         e[y].push_back((node){x,r,c});
68     } 
69     if(spfa(num))   
70         cout<<"YES"<<endl;
71     else    
72         cout<<"NO"<<endl;   
73 
74     return 0;
75 }