(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


[align=left]Problem Description[/align]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.
 
[align=left]Input[/align]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. 
 
[align=left]Output[/align]For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No". 
 
[align=left]Sample Input[/align]

3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0 
[align=left]Sample Output[/align]

Case 1: Yes
Case 2: No 
[align=left]Source[/align]University of Ulm Local Contest 1996 
[align=left]Recommend[/align]Eddy   |   We have carefully selected several similar problems for you:  1142 1385 2112 1598 2923 

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;
}