(floyd 1.1)hdu 1217 Arbitrage(使用floyd来求最长路——判断是否存在一种货币，经过一个兑换回路以后>=1单元)

题目：

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5052    Accepted Submission(s): 2318

Problem Description

Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent. 

Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

 

Input

The input file will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n. 

 

Output

For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No". 

 

Sample Input

3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar0

 

Sample Output

Case 1: YesCase 2: No

 

Source

University of Ulm Local Contest 1996

 

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题目分析：

               使用floyd来求最长路。这道题需要注意以下的一些地方。

1）文字索引需要转化成数字索引。因为数组里面是数字索引。

string huobi_name;int i;for(i = 0 ; i < n ; ++i){//***将文字索引转换成数字索引.因为数组中用的是数字索引cin >> huobi_name;money[huobi_name] = huobi_type++;}

2）求最长路时的floyd的初始化方法的写法。

/** * 这道题中floyd的初始化. * 以前的写法如上提所示. * 这里讲一下区别。以前是求最短路径,是一个由大到小的紧缩过程。 * 而这道题抽象以下其实是求最长路径,是一个由小到大的放大的过程 */void initial() {int i, j;for (i = 1; i <= n; ++i) {for (j = 1; j <= n; ++j) {e[i][j] = 0;//所有东西一开始都初始化位0}}}

3）求最长路时floyd函数的放缩的写法。其实就是把 > 换成  < 即可。

 

/** *floyd算法 */void floyd() {int i, j, k;for (k = 1; k <= n; ++k) {//遍历所有的中间点for (i = 1; i <= n; ++i) {//遍历所有的起点for (j = 1; j <= n; ++j) {//遍历所有的终点//*****floyd求最长路的写法if (e[i][j] < e[i][k] * e[k][j]) {//如果当前i-->j的距离   小于   i-->k--->j的距离之和e[i][j] = e[i][k] * e[k][j];//更新从i--->j的最短路径}}}}}

这道题的代码如下：

/* * hdu1217.cpp * *  Created on: 2015年3月27日 *      Author: Administrator */#include <iostream>#include <cstdio>#include <map>using namespace std;const int maxn = 31;double e[maxn][maxn];int n;const int inf = 99999999;/* * void initial() {int i, j;for (i = 1; i <= n; ++i) {for (j = 1; j <= n; ++j) {if (i == j) {e[i][j] = 0;} else {e[i][j] = inf;}}}} * *//** * 这道题中floyd的初始化. * 以前的写法如上提所示. * 这里讲一下区别。以前是求最短路径,是一个由大到小的紧缩过程。 * 而这道题抽象以下其实是求最长路径,是一个由小到大的放大的过程 */void initial() {int i, j;for (i = 1; i <= n; ++i) {for (j = 1; j <= n; ++j) {e[i][j] = 0;//所有东西一开始都初始化位0}}}/** *floyd算法 */void floyd() {int i, j, k;for (k = 1; k <= n; ++k) {//遍历所有的中间点for (i = 1; i <= n; ++i) {//遍历所有的起点for (j = 1; j <= n; ++j) {//遍历所有的终点//*****floyd求最长路的写法if (e[i][j] < e[i][k] * e[k][j]) {//如果当前i-->j的距离   小于   i-->k--->j的距离之和e[i][j] = e[i][k] * e[k][j];//更新从i--->j的最短路径}}}}}int main(){int cnt = 1;while(scanf("%d",&n)!=EOF,n){map<string,int> money;int huobi_type = 1;string huobi_name;int i;for(i = 0 ; i < n ; ++i){//***将文字索引转换成数字索引.因为数组中用的是数字索引cin >> huobi_name;money[huobi_name] = huobi_type++;}initial();int m;scanf("%d",&m);while(m--){string from;string to;double change;cin >> from >> change >> to;e[money[from]][money[to]] = change;}floyd();bool flag = false;for(i = 1 ; i <= n ; ++i){//遍历所有的货币if(e[i][i] > 1){//看是否存在一种货币经过一个兑换回路以后 >= 1个单元//如果存在flag = true;//则将flag标记为true。break;//跳出循环}}//输出最后的结果if(flag == true){printf("Case %d: Yes\n",cnt++);}else{printf("Case %d: No\n",cnt++);}}return 0;}