Zhao70's Blog

POJ1860 Currency Exchange

发表于 | 分类于

题目大意

POJ1860

给定一个带有点权和边权的图, 求是否存在正权回路.

思路

由于Bellman-Ford算法可以求图中是否存在负权回路. 因此可以将 dis数组的初始状态置为 0, 求是否存在正权回路.

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
#include <iostream>
#include <vector>
using namespace std;
struct Edge{
    Edge(int a_, int b_, double r_, double c_)
        : a(a_), b(b_), rate(r_), commi(c_){}
    int a, b;
    double rate, commi;
};
const int arrSize = 210;
int main(){
    int nEdge, nVertex, amoCur;
    double vCur;
    cin >> nVertex >> nEdge >> amoCur >> vCur;
    vector<Edge> edge;
    for(int i = 0; i < nEdge; ++i){
        int a, b;
        cin >> a >> b;
        double r, c;
        cin >> r >> c;
        edge.push_back(Edge(a, b, r, c));
        cin >> r >> c;
        edge.push_back(Edge(b, a, r, c));
    }
    double dis[arrSize];
    for(int i = 0; i < nEdge * 2; i++)
        dis[i] = 0;
    dis[amoCur] = vCur;
    for(int i = 0; i <= nVertex; i++){
        bool flag = true;
        for(int j = 0; j < edge.size(); j++){
            if(dis[edge[j].b] < (dis[edge[j].a] - edge[j].commi) * edge[j].rate){
                dis[edge[j].b] = (dis[edge[j].a] - edge[j].commi) * edge[j].rate;
                flag = false;
            }
        }
        if(flag)
            break;
        if(i == nVertex){
            cout << "YES" << endl;
            return 0;
        }
    }
    cout << "NO" << endl;
    return 0;
}