HDU6301 SET集合的应用 贪心

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<j≤r), ai≠aj

holds.
Chiaki would like to find a lexicographically minimal array which meets the facts.

InputThere 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.
OutputFor 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

这个题目好就好在贪心的时候用两个指针维护，时间复杂度还是不超过O(n)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>

#define LL long long
using namespace std;

int n,m;
int answer[100010];
struct node
{
    int l,r;
}a[100010];
int cmp(node x,node y)
{
    if(x.l==y.l)
        return x.r>y.r;
    return x.l<y.l;
}
set<int>t;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        t.clear();
        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; i++){
            t.insert(i);
            answer[i] = 0;
        }
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d",&a[i].l,&a[i].r);
        }
        sort(a,a+m,cmp);
        int l = 1,r = 1;
        for(int i = 0; i < m; i++)
        {
            for(int j = l; j < a[i].l; j++)
            {
                if(answer[j]!=0)
                {
                    t.insert(answer[j]);
                }
                l++;
            }
            if(r<l) r = l;
            for(int j = r; j <= a[i].r; j++)
            {
                if(answer[j]==0){
                    answer[j] = *t.begin();
                    t.erase(t.begin());
                }
                r++;
            }
        }
        for(int i = 1; i < n; i++)
            if(answer[i]==0)
                printf("1 ");
            else
                printf("%d ",answer[i]);
        if(answer[n]==0)
            printf("1
");
        else
            printf("%d
",answer[n]);
    }
    return 0;
}