Distinct Values HDU - 6301（贪心）HDU 7.23多校网络赛

Distinct Values HDU - 6301

先感谢下在B站直播讲题的清华大佬。ˋ( ° ▽、° )

题目：

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≤10e5) – 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 10e6.

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

题面意思：

有t组样例，有个n长度的数组和m组限制。每一组限制输入为 l 和 r ，在数组 l 下标到 r 下标的范围内不能有重复数字。

思路：

算法是贪心。
每次输入l 和 r 时，把r 下标的数组数据改为 l。m组限制输入完毕后，从设置的下标数组从后往前比较，倒数第二个和倒数第一个比较，一直遍历到第一个。这样就能把每个最后要限制的不同的范围的数据改为l 的下标。然后利用set默认从小到大排序的便利，先初始化set（从1到n），每个范围内开始遍历每次有可以用的数据就从set中提出并erase去除。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
using namespace std;

const int SIZE = 1e5+5;
int pre[SIZE];
int arr[SIZE];
set<int> s;
int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        int m, n, l, r;
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++) {
            pre[i] = i;
        }
        for(int i = 0; i < m; i++) {
            scanf("%d%d", &l, &r);
            pre[r] = min(l, pre[r]);
        }
        for(int i = n-1; i >= 1; i--) {
            pre[i] = min(pre[i], pre[i+1]);
        }
        s.clear();
        for(int i = 1; i <= n; i++) {
            s.insert(i);
        }
        int js = 1;
        for(int i = 1; i <= n; i++) {
            while(js < pre[i]) {
                s.insert(arr[js]);
                js++;
            }
            arr[i] = *s.begin();
            s.erase(arr[i]);
        }
        for(int i = 1; i <= n; i++) {
            printf("%d", arr[i]);
            printf("%c", i == n ? '\n' : ' ');
        }
    }
    return 0;
}