【哈希】POJ 3640 Conformity

Conformity

解题思路

原来POJ用STL会T…

对每个组合，读进来之后先把 5 个数排序，拼成 15 位的数，之后去统计每个数出现的次数。

参考代码

#include<stdio.h>
#include<iostream>
#include<vector>
#include<cstring>
#include<cstdio>
#include<climits>
#include<cmath>
#include<algorithm>
#include<queue>
#include<deque>
#include<map>
#include<set>
#include<stack>
#define LL long long
#define LOCAL
const int maxn = 1e4 + 10;

using namespace std;

int readint() {
    
    
    int x; scanf("%d", &x); return x;
}

int tmp[10];
LL num[maxn];
int cnt[maxn];

int main() {
    
    
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
//   freopen("output.txt", "w", stdout);
#endif
    int n;
    while (~scanf("%d", &n) && n) {
    
    
        for(int i = 0; i < n; i++) {
    
    
            for(int k = 0; k < 5; k++){
    
    
                cnt[k] = readint();
            }
            sort(cnt, cnt + 5);
            LL val = 0;
            for(int j = 0; j < 5; j++) {
    
    
                val = val * 1000 + cnt[j];
            }
            num[i] = val;
        }
        sort(num, num + n);
        int k = 1, ccnt = 0;
        for(int i = 0; i + 1 < n; i++) {
    
    
            if (num[i] == num[i + 1]) k++;
            else {
    
    
                cnt[ccnt++] = k;
                k = 1;
            }
        }
        cnt[ccnt++] = k;
        int ansm = 0, ans = 0;
        for(int i = 0; i < ccnt; i++) {
    
    
            if (cnt[i] == ansm) ans++;
            if (cnt[i] > ansm) {
    
    
                ansm = cnt[i];
                ans = 1;
            }
        }
        printf("%d\n", ans * ansm);
    }
    return 0;
}