【题目链接】

Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Chiaki has an array of n positive integers. You are told some facts about the array: for every two elements ai and aj in the subarray al..r (l≤i

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers n and m (1≤n,m≤105) – the length of the array and the number of facts. Each of the next m lines contains two integers li and ri (1≤li≤ri≤n).

It is guaranteed that neither the sum of all n nor the sum of all m exceeds 106.

Output

For each test case, output n integers denoting the lexicographically minimal array. Integers should be separated by a single space, and no extra spaces are allowed at the end of lines.

Sample Input

3
2 1
1 2
4 2
1 2
3 4
5 2
1 3
2 4

Sample Output

1 2
1 2 1 2
1 2 3 1 1

Source

2018 Multi-University Training Contest 1

先初始化set，que，ans数组，然后对m组数据预处理que[a] = max(que[a], b)。
当遍历到i时，set里面存的是ans[i]之后没有出现的数字，所以在每次i++后，都要将ans[i-1]重新放入到ans数组中，
当遍历到i时，最少应该输出que[i]个数字，若此时输出的数字小于que[i]，则从set从取出que[i]-i个元素放入ans数组中，反之，不进行任何操作。

下面给一组案例

1
5 2
1 3
2 4

#include <iostream>
#include <set>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int n, m, a, b;
        scanf("%d%d", &n, &m);
        int que[100005];
        for (int i = 1; i <= n; i++)
            que[i] = i;
        for (int i = 0; i < m; i++)
        {
            scanf("%d%d", &a, &b);
            que[a] = max(que[a], b);
        }
        int ans[100005];
        set<int>s;
        for (int i = 1; i <= n; i++)
            s.insert(i);
        int output = 1, input = 1;
        for (int i = 1; i <= n; i++)
        {
            if (i != 1)
                s.insert(ans[i - 1]);
            while (output <= que[i])
            {
                ans[output] = *s.begin();
                s.erase(ans[output++]);
            }
        }
        for (int i = 1; i <= n; i++)
            printf("%d%c", ans[i], i == n ? '\n' : ' ');
    }
}