Educational Codeforces Round 6 620C Pearls in a Row(stl)

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本文介绍了一种将连续排列的珍珠分成尽可能多的相同类型的连续段落的方法，并提供了实现该策略的代码示例。

There are n pearls in a row. Let's enumerate them with integers from 1 to n from the left to the right. The pearl number i has the type ai.

Let's call a sequence of consecutive pearls a segment. Let's call a segment good if it contains two pearls of the same type.

Split the row of the pearls to the maximal number of good segments. Note that each pearl should appear in exactly one segment of the partition.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printfinstead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains integer n (1 ≤ n ≤ 3·105) — the number of pearls in a row.

The second line contains n integers ai (1 ≤ ai ≤ 109) – the type of the i-th pearl.

Output

On the first line print integer k — the maximal number of segments in a partition of the row.

Each of the next k lines should contain two integers lj, rj (1 ≤ lj ≤ rj ≤ n) — the number of the leftmost and the rightmost pearls in the j-th segment.

Note you should print the correct partition of the row of the pearls, so each pearl should be in exactly one segment and all segments should contain two pearls of the same type.

If there are several optimal solutions print any of them. You can print the segments in any order.

If there are no correct partitions of the row print the number "-1".

 
       Sample test(s)
      
         input
        
5
1 2 3 4 1
 
         output
        
1
1 5
 
         input
        
5
1 2 3 4 5
 
         output
        
-1
 
         input
        
7
1 2 1 3 1 2 1
 
         output
        
2
1 3
4 7

题目链接:点击打开链接

给出一个序列, 一段子序列含有两个相同的数字称为happy segment, 要求分为最多的happy segment, 输出每段序列起点终点.

考虑存储数据的数据结构, 使用set解决此问题, 遇到重复的元素则存储位置且清空set, 同时要满足第一段序列的起点为1且最后一段序

列的终点为n 即可.

AC 代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 3e5 + 5;
int n, pos[MAXN], k;
set<int> s;
int main(int argc, char const *argv[])
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		int x;
		scanf("%d", &x);
		if(s.count(x) == 0) s.insert(x);
		else {
			pos[k++] = i;
			s.clear();
		}
	}
	if(k == 0) printf("-1\n");
	else {
		pos[k - 1] = n;
		printf("%d\n", k);
		printf("1 %d\n", pos[0]);
		for(int i = 0; i < k - 1; ++i)
			printf("%d %d\n", pos[i] + 1, pos[i + 1]);
	}
	return 0;
}