B - Amr and The Large Array

题目链接：http://codeforces.com/problemset/problem/558/B

 

题意：

一个序列的美丽程度与其中某个数重复次数的最大值有关。求最短的子序列（连续的一段）使得其美丽程度与原序列相等。

 

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <set>
#include <queue>
#include <stdbool.h>

#define LL long long
using namespace std;
const int maxn = 1e5 + 10;


struct Node{
    int val;
    int pos;
}arr[maxn];

bool cmp(Node a,Node b){
    if (a.val != b.val){
        return a.val < b.val;
    }
    else{
        return a.pos < b.pos;
    }
}


int main(){
    int n;
    scanf("%d",&n);
    for (int i=1;i<=n;i++){
        scanf("%d",&arr[i].val);
        arr[i].pos = i;
    }
    sort(arr+1,arr+1+n,cmp);
    int ansl,ansr;
    int start = 1;
    int len = 0,cnt = 1;
    for (int i=2;i<=n;i++){
        if (arr[i].val == arr[i-1].val){
            cnt++;
        }
        else{
            if (cnt > len){
                ansl = arr[start].pos;
                ansr = arr[i-1].pos;
                len = cnt;
            }
            else if (cnt == len){
                if (arr[i-1].pos-arr[start].pos+1 < ansr-ansl+1){
                    ansl = arr[start].pos;
                    ansr = arr[i-1].pos;
                }
            }
            start = i;
            cnt = 1;
        }
    }
    if (cnt > len){
        ansl = arr[start].pos;
        ansr = arr[n].pos;
    }
    else if (cnt == len){
        if (arr[n].pos-arr[start].pos+1 < ansr-ansl+1){
            ansl = arr[start].pos;
            ansr = arr[n].pos;
        }
    }
    printf("%d %d\n",ansl,ansr);
    return 0;
}