Hangzhou Electric 2094--Generation Champion (topological sort)

Produce Champions

Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 11549 Accepted Submission (s): 5360

Problem description A group of people, playing table tennis game, 22 catch to kill, every two people play a maximum of a game.
The rules of the game are as follows:
If a defeated B,b and defeated C, and A and C did not play, then it is decided that a must be able to defeat C.
If a defeated B,b and defeated C, and C defeated A, then a, B, C Three will not be the champion.
According to this rule, there is no need to cycle a contest, perhaps to determine the championship. Your task is to face a group of contestants, after a number of killings, to determine whether it has actually produced a championship.

Input inputs contain a number of contestants, each of which starts with an integer n (n<1000) followed by N against the competitor, and the result is a pair of player names (one space between them), the former defeating the latter. If n is 0, it indicates the end of the input.

Output for each player group, if you determine that a winner is generated, export "Yes" in one line, otherwise output "No" in a row.

Sample Input3 Alice Bob Smith John Alice Smith5 a cc D d e b ea d 0

Sample Outputyes No

Authorqianneng

Source greets new semester-super easy version

Recommendlcy | We have carefully selected several similar problems for you:2093 1811 2086 2647 2090 RE: It is said that a winner is one who has not lost.

1#include <map>2#include <cstdio>3#include <cstring>4#include <iostream>5 using namespacestd;6map<string,int>acc;7 intvis[1010], pic[1010][1010], indegree[1010];8 BOOLFlag;9 intN, K, top;Ten voidDeal () One { Amemset (Vis,0,sizeof(Vis)); -memset (pic,0,sizeof(pic)); -memset (Indegree,0,sizeof(Indegree)); theAcc.clear ();stringA, B; K =0; -          for(inti =0; I < n; i++) -         { -Cin >> a >>b; +             if(!acc[a]) acc[a] = + +K; -             if(!acc[b]) acc[b] = + +K; +             if(Pic[acc[a]][acc[b]] = =0) A             { atPIC[ACC[A]][ACC[B]] =1; -indegree[acc[b]]++;  -             }     -         } -         intTotal =0;  -          for(inti =1; I <= K; i++) in             if(!Indegree[i]) -total++; to         if(Total! =1) +Flag =false; - } the voidTsort () * { $top =0;inti;Panax Notoginseng      while(Top <k) -     { the          for(i =1; I <= K; i++) +             if(!vis[i] &&!Indegree[i]) A                  Break; theVis[i] =1; +top++; -          for(intj =1; J <= K; J + +) $             if(Pic[i][j]) $indegree[j]--; -     } - }  the  - /*void Tsort () {//wa: There is no guarantee that there is only one person who has not lost. Wuyi int m; the For (int j = 0; J < K; J + +) { - m =-1; Wu for (int i = 1; I <= K; i++) - if (!indegree[i]) { About m = i; $ Break ; -             } - if (m = =-1) { - flag = false; A Break ; +             } the indegree[m]--; - for (int i = 1; I <= K; i++) { $ if (Pic[m][i]) the indegree[i]--; the             } the     } the } */ -  in intMain () the { the      while(~SCANF ("%d", &N), N) About     { theFlag =true; the Deal (); the Tsort (); +         if(Top <k)//seems to be a judgment ring.  -Flag =false; the         if(flag)Bayiprintf"yes\n"); the         Else theprintf"no\n"); -     } -     return 0; the}

Hangzhou Electric 2094--Generation Champion (topological sort)