POJ T2240 Arbitrage（货币流通问题）

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                                      POJ T2240 Arbitrage

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题目思路：

    本题跟POJ 1806思路一样，判断正环，只是要给货币种类编号。用STL的话可以用map 映射，鄙人用的是普通遍历匹配— . —

 

代码:

#include<cstdio>
#include<iostream>
#include<algorithm>
#define maxn 35
using namespace std;

double dis[maxn];
int n,m;
string name[maxn];

struct Edge{
	int u,v;
	double r;
}edge[maxn*maxn];

int find(string s){
	for(int i = 1; i <= n; i++){
		if(s == name[i]) return i;  //给货币编号 
	}
}

void Init(){
	for(int i = 1; i <= n; i++){
		dis[i] = 0;    //初始化为0 
	}
}

bool Bellman_ford(int start){
	dis[start] = 1;
	bool flag;
	for(int i = 1; i < n; i++){
		flag = true;
	    for(int j = 1; j <= m; j++){
	        if(dis[edge[j].v] < dis[edge[j].u] * edge[j].r){
	        	dis[edge[j].v] = dis[edge[j].u] * edge[j].r;
	        	flag = false;
	        }
	    }
	    if(flag) break;
	}
	for(int j = 1; j <= m; j++)
	    if(dis[edge[j].v] < dis[edge[j].u] * edge[j].r)
	        return true;
	return false;
}


int main(){
	string s1,s2;
	int kcase = 0;
	while(scanf("%d",&n)&&n){
		Init();
		for(int i = 1; i <= n ;i++)
		    cin >> name[i];
		scanf("%d",&m);
		for(int i = 1; i <= m; i++){
			cin >> s1;
			scanf("%lf",&edge[i].r);
			cin >> s2;
			edge[i].u = find(s1);
			edge[i].v = find(s2);
		}
		printf("Case %d: ",++kcase);
		if(Bellman_ford(1)) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}