[UVA][Trie] 11286 - Conformity＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2012-04-01 20:28:48| 人1,206| 回0 | 上一篇 | 下一篇

[UVA][Trie] 11286 - Conformity

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Problem B: Conformity

Frosh commencing their studies at Waterloo have diverse interests,as evidenced by their desire to take various combinations of courses fromamong those available.

University administrators are uncomfortable with this situation, andtherefore wish to offer a conformity prize to frosh who choosethe most popular combination of courses. How many frosh will winthe prize?

The input consists of several test cases followed by a line containing0. Each test case begins with an integer 1 ≤ n ≤ 10000, the number offrosh. For each frosh, a line follows containing the course numbersof five distinct courses selected by the frosh. Each course number isan integer between 100 and 499.

The popularity of a combination is the number of frosh selectingexactly the same combination of courses. A combination of coursesis considered most popular if no other combination has higherpopularity. For each line of input, you should output a single linegiving the total number of students taking some combination of courses thatis most popular.

Sample Input

3
100 101 102 103 488
100 200 300 101 102
103 102 101 488 100
3
200 202 204 206 208
123 234 345 456 321
100 200 300 400 444
0

Output for Sample Input

2
3

最受迎的合可能有很多, 出最受迎合的人

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct Trie {
 int n;
 int link[10];
}Node[200000];
int TrieSize;
void insertTrie(const char* str) {
 static int i, idx;
 idx = 0;
 for(i = 0; str[i]; i++) {
 if(Node[idx].link[str[i]-'0'] == 0) {
 TrieSize++;
 memset(&Node[TrieSize], 0, sizeof(Node[0]));
 Node[idx].link[str[i]-'0'] = TrieSize;
 }
 idx = Node[idx].link[str[i]-'0'];
 }
 Node[idx].n ++;
}
int max, maxNum;
void findAns(int now) {
 if(Node[now].n != 0) {
 if(max == Node[now].n) {
 maxNum += Node[now].n;
 } else if(max < Node[now].n){
 max = Node[now].n;
 maxNum = max;
 }
 }
 int i;
 for(i = 0; i < 10; i++) {
 if(Node[now].link[i] != 0) {
 findAns(Node[now].link[i]);
 }
 }
}
int cmp(const void *i, const void *j) {
 return *(int *)i - *(int *)j;
}
int main() {
 int n, A[5];
 char buf[20];
 while(scanf("%d", &n) == 1 && n) {
 TrieSize = 0;
 memset(&Node[0], 0, sizeof(Node[0]));
 while(n--) {
 scanf("%d %d %d %d %d", &A[0], &A[1], &A[2], &A[3], &A[4]);
 qsort(A, 5, sizeof(int), cmp);
 sprintf(buf, "%d%d%d%d%d", A[0], A[1], A[2], A[3], A[4]);
 insertTrie(buf);
 }
 max = 0, maxNum = 0;
 findAns(0);
 printf("%d\n", maxNum);
 }
 return 0;
}