hdu Distinct Values

                Distinct Values

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 36   Accepted Submission(s) : 17

Problem Description

Chiaki has an array of $n$ positive integers. You are told some facts about the array: for every two elements $a_i$ and $a_j$ in the subarray $a_{l..r}$ ($l \le i < j \le r$), $a_i \ne a_j$ holds.
Chiaki would like to find a lexicographically minimal array which meets the facts.

 

 

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 $m$ ($1 \le n, m \le 10^5$) -- the length of the array and the number of facts. Each of the next $m$ lines contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$). It is guaranteed that neither the sum of all $n$ nor the sum of all $m$ exceeds $10^6$.

 

 

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

 

这道题的意思是组成最小的数，需要用到STL排序。

#include<set>
#include<bits/stdc++.h>
const int N = 5e5 + 10;
const int INF=0x3f3f3f3;
using namespace std;
int a[N],pre[N];//用来记录下标，代表着区间里的元素不能重复；
int main()
{
    int i,m,n;
    int t,qi,hou;//qi代表区间的前边界，hou代表区间的后边界；
    scanf("%d",&t);
    while(t--)
    {

        scanf("%d%d",&m,&n);
        for(i=1; i<=m; i++)
        {
            pre[i]=i;
        }
        for(i=1; i<=n; i++)
        {
            scanf("%d%d",&qi,&hou);
            pre[hou]=min(pre[hou],qi);
        }
        for(i=m-1; i>=1; i--)
        {
            pre[i]=min(pre[i],pre[i+1]);把同意区间的元素都标记为区间的前元素，代表同属一个集合，不能重复
        }
        set<int> s;
        int p=1;
        for(i=1; i<=m; i++)
        {
            s.insert(i);
        }
        for(i=1; i<=m; i++)
        {
            while(p<pre[i])//p代表下表，如果p小于区间的前元素，则不属于这个区间，讲p对应的元素传到s里面
            {
                s.insert(a[p]);
                p++;
            }
            a[i]=*s.begin();
            s.erase(a[i]);
        }
        for(i=1; i<=m; i++)
        {
            if(i==1)
                printf("%d",a[i]);
            else
                printf(" %d",a[i]);
        }
        puts("");
    }
    return 0;
}