(Weird Floyd & self-ring) MZ training 2014 #15 E question (poj 2240)

Have you ever seen such a strange Floyd?

First.

[Description]

There is a difference in the exchange rate of the currency, for example, assume that 1 dollar buys 0.5, 1 pound buys 10 francs, and 1 French francs buys 0.21 dollars. Then, by currency conversion, a smart trader can buy 0.5*10*0.21 = 1.05 USD from 1 USD, making a profit of 5%.

Your task is to write a program that uses a currency exchange rate list as the input, and then determine whether or not you can make a profit.

[Intput]

The input contains multiple groups of test data. The first row of each group contains N (1 <= n <= 30), which indicates that there are n medium currencies and the next n rows, each row is a currency name with no space in the name. The next row contains an integer m, followed by M. Each row contains three parts: Si, rij, SJ, it indicates that the exchange rate of Si to SJ is rij. Each group of data is separated by an empty row. When n = 0, the input ends.

 

[Output]

Output "case I: YES" or "case I: No" for each group of data indicates whether group I data is feasible.

 

[Sample input]

3

Usdollar

Britishpound

Frenchfranc

3

Usdollar 0.5 briishpound

Britishpound 10.0 frenchfranc

Frenchfrankc 0.21 usdollar

 

3

Usdollar

Britishpound

Frenchfranc

6

Usdollar 0.5 briishpound

Usdollar 4.9 frenchfranc

Britishpound 10.0 frenchfranc

Britishpound 1.99 usdollar

Frenchfranc 0.09 britishpound

Frenchfrankc 0.19 usdollar

 

0

[Sample onput]

Case 1: Yes

Case 2: No

At first glance, it looks like a mathematical problem. Actually, it's Graph Theory (I believe everyone can see it ).

Consider exchange rates as edges and currencies as points.

Floyd.

<span style="color:#660000;"><span style="font-size:18px;">for(int k=1;k<=n;k++)</span><span style="font-size:14px;"></span><span style="font-size:18px;">for(int i=1;i<=n;i++)for(int j=1;j<=n;j++){if(i==j || i==k || j==k) continue;if(dis[i][j]<dis[i][k]*dis[k][j]){dis[i][j]=dis[i][k]*dis[k][j];}}</span></span>

Pay special attention to the transfer equation of this question: Of course, the exchange rate or something must be multiplied.

Note: This question cannot be ruled out from the self-loop (my subconscious Code allows me to adjust for a few hours), that is, I changed my own exchange rate more than 1 (this Nima is unscientific !)

<Span style = "color: # 3366ff; Background-color: RGB (102,255,153 ); "> # include <cstdio> # include <iostream> # include <cstring> # include <cstdlib> # include <algorithm> # include <cmath> using namespace STD; int N, m; double dis [35] [35]; char name [35] [3110]; char TMP [3110]; int ansnum = 0; int main () {While (1) {scanf ("% d", & N); If (n = 0) break; memset (name, 0, sizeof (name); For (INT I = 1; I <= N; I ++) for (Int J = 1; j <= N; j ++) dis [I] [J] = 0.0; for (INT I = 1; I <= N; I ++) scanf ("% s", name [I]); scanf ("% d", & M ); int U, V; double W; For (INT I = 1; I <= m; I ++) {memset (TMP, 0, sizeof (TMP )); scanf ("% s", TMP); For (Int J = 1; j <= N; j ++) {If (strcmp (TMP, name [J]) = 0) {u = J; break;} scanf ("% lf", & W); memset (TMP, 0, sizeof (TMP )); scanf ("% s", TMP); For (Int J = 1; j <= N; j ++) {If (strcmp (TMP, name [J]) = 0) {v = J; break;} dis [u] [v] = max (DIS [u] [v], W );} for (int K = 1; k <= N; k ++) for (INT I = 1; I <= N; I ++) for (Int J = 1; j <= N; j ++) {if (I = j | I = k | j = k) continue; if (DIS [I] [J] <dis [I] [k] * Dis [k] [J]) {dis [I] [J] = dis [I] [k] * Dis [k] [J] ;}} int OK = 0; For (INT I = 1; I <= N; I ++) {for (Int J = 1; j <= N; j ++) {// if (I = J) continue; this statement must not contain if (DIS [I] [J] * Dis [J] [I]> 1.0) {OK = 1 ;}if (OK) break ;} if (OK) break;} printf ("case % d:", ++ ansnum); If (OK) printf ("Yes \ n "); else printf ("NO \ n");} return 0 ;}</span>

(Weird Floyd & self-ring) MZ training 2014 #15 E question (poj 2240)