Description
Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency.
For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR.
You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real R AB, C AB, R BA and C BA - exchange rates and commissions when exchanging A to B and B to A respectively.
Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations.Input
The first line of the input contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1<=S<=N<=100, 1<=M<=100, V is real number, 0<=V<=10 3.
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10 -2<=rate<=10 2, 0<=commission<=10 2.
Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 10 4.Output
If Nick can increase his wealth, output YES, in other case output NO to the output file.Sample Input
3 2 1 20.0 1 2 1.00 1.00 1.00 1.00 2 3 1.10 1.00 1.10 1.00Sample Output
YES
更多的样例：链接
思路
我们把每种货币看成图的顶点，而每个交换站点实现一对货币的交换，可认为增加一个交换站点就是增加两个顶点间的一个环路。从样例中可以知道，如果想要让NICK的资金增加，那么就需要存在一个权值为正的环路，使得货币总价值能够无限上升。
所以现在的问题变成了如何判断图中是否有正环。由于货币交换后总价值可能降低，也就是说可能存在负权边，那么 Dijkstra 就不适用了，应该采用能判断负环的 Bellman_ford ，还有它的优化算法 SPFA 。由于需要判断图中是否存在能使货币总价值无限上升的正环，那么松弛边的代码也就不是原来求最短路的松弛操作了，需要变成了 dis[v] < (dis[u] - C) * R 。改了松弛操作，原本判断负环的操作也就对应的变成了判断正环。
AC 代码如下：
1. BFS 判断货币交换后是否增值。依据：货币交换相当于增加环路，那么从 S 出发就总能回到 S，通过 S 的值来判断其是否增值。
#include<iostream> #include<algorithm> #include<queue> #include<vector> #include<cstring> using namespace std; const int INF = 0x3f3f3f3f; const int MAXN = 105; int N, M; double V; struct Edge{ int to; double R, C; }; vector<Edge> G[MAXN]; //G[i]代表从i出发的边,vector存储边 void addEdge (int u, int v, double r, double c) { Edge tmp; tmp.to = v; tmp.R = r; tmp.C = c; G[u].push_back(tmp); } double dis[MAXN]; bool inQueue[MAXN]; //spfa算法判断dis[s]是否可能大于V (实质是BFS) bool spfa (int s) { for (int i = 1; i <= N; i++) {dis[i] = 0.0; inQueue[i] = false; } dis[s] = V; queue<int> q; q.push(s); inQueue[s] = true; while (!q.empty()) { int u = q.front(); q.pop(); inQueue[u] = false; for (int j = 0; j < G[u].size(); j++) { int v = G[u][j].to; double r = G[u][j].R, c = G[u][j].C; if (dis[v] < (dis[u] - c)*r ) { dis[v] = (dis[u] - c)*r; if (!inQueue[v]) { q.push (v); inQueue[v] = true; } } } if (dis[s] > V) { return true; } } return false; } int main(void) { int S; while (cin >> N >> M >> S >> V) { for (int i = 1; i <= N; i++) G[i].clear(); for (int i = 1; i <= M; i++) { int u, v; double R1, C1, R2, C2; cin >> u >> v >> R1 >> C1 >> R2 >> C2; addEdge(u, v, R1, C1); addEdge(v, u, R2, C2); } //spfa算法判断dis[s]是否可能大于V if (spfa(S)) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }
2.Bellman_ford 判断是否存在正环。
#include<iostream> #include<algorithm> #include<queue> #include<vector> #include<cstring> using namespace std; const int INF = 0x3f3f3f3f; const int MAXN = 105; int N, M; double V; struct Edge{ int to; double R, C; Edge(int v, double r, double c) : to(v), R(r), C(c) {} }; vector<Edge> G[MAXN]; //G[i]代表从i出发的边,vector存储边 void addEdge (int u, int v, double r, double c) { //Edge tmp; tmp.to = v; tmp.R = r; tmp.C = c; //G[u].push_back(tmp); G[u].push_back(Edge(v, r, c)); } double dis[MAXN]; bool bellman_ford (int s) { for (int i = 1; i <= N; i++) dis[i] = 0.0; dis[s] = V; bool relaxed = false; //哨兵 //最多更新N-1次dis数组 for (int i = 1; i <= N-1; i++) { relaxed = false; //遍历M条边 for (int u = 1; u <= N; u++) { for (int j = 0; j < G[u].size(); j++) { int v = G[u][j].to; double r = G[u][j].R, c = G[u][j].C; if (dis[v] < (dis[u] - c)*r ) { dis[v] = (dis[u] - c)*r; relaxed = true; } } } if (!relaxed) break; } //判断是否存在正环 for (int u = 1; u <= N; u++) { for (int j = 0; j < G[u].size(); j++) { int v = G[u][j].to; double r = G[u][j].R, c = G[u][j].C; //如果还有边可以松弛，说明存在正环 if (dis[v] < (dis[u] - c)*r ) { return false; } } } return true; //返回true说明图不包含正环 //或者用spfa算法直接判断dis[s]是否大于V } int main(void) { int S; while (cin >> N >> M >> S >> V) { for (int i = 1; i <= N; i++) G[i].clear(); for (int i = 1; i <= M; i++) { int u, v; double R1, C1, R2, C2; cin >> u >> v >> R1 >> C1 >> R2 >> C2; addEdge(u, v, R1, C1); addEdge(v, u, R2, C2); } //bellman_ford算法判断是否存在正环 if (!bellman_ford(S)) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }
3.SPFA 判断是否存在正环。 SPFA算法的模板及分析在我的另一篇博客里：链接
#include<iostream> #include<algorithm> #include<queue> #include<vector> #include<cstring> using namespace std; const int INF = 0x3f3f3f3f; const int MAXN = 105; int N, M; double V; struct Edge{ int to; double R, C; }; vector<Edge> G[MAXN]; //G[i]代表从i出发的边,vector存储边 void addEdge (int u, int v, double r, double c) { Edge tmp; tmp.to = v; tmp.R = r; tmp.C = c; G[u].push_back(tmp); } double dis[MAXN]; bool inQueue[MAXN]; int cnt[MAXN]; //spfa判断是否存在正环 bool spfa (int s) { memset (cnt, 0, sizeof(cnt)); memset (dis, 0.0, sizeof(dis)); memset (inQueue, false, sizeof(inQueue)); dis[s] = V; queue<int> q; q.push(s); inQueue[s] = true; while (!q.empty()) { int u = q.front(); q.pop(); inQueue[u] = false; for (int j = 0; j < G[u].size(); j++) { int v = G[u][j].to; double r = G[u][j].R, c = G[u][j].C; if (dis[v] < (dis[u] - c)*r ) { dis[v] = (dis[u] - c)*r; if (!inQueue[v]) { q.push (v); inQueue[v] = true; if (++cnt[v] > N) return false; } } } } return true; //返回true说明图中不存在正环 } int main(void) { int S; while (cin >> N >> M >> S >> V) { for (int i = 1; i <= N; i++) G[i].clear(); for (int i = 1; i <= M; i++) { int u, v; double R1, C1, R2, C2; cin >> u >> v >> R1 >> C1 >> R2 >> C2; addEdge(u, v, R1, C1); addEdge(v, u, R2, C2); } //spfa算法判断是否存在正环 if (!spfa(S)) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }