HDU多校（Distinct Values）

 

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<j≤r ), ai≠aj 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≤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

 

这回的航电多校有多个签到题 好评 

 

  这题就用L,R 两个指针维护这个ans序列

　

while(L < qu[i].l) {
     st.insert(ans[L]);
     L++;
}

这些值可以重新插入

while(R < qu[i].r) {
if (R < qu[i].l-1) R++;
else {
       R++;
      ans[R] = *st.begin();
      st.erase(st.begin());
      }
}

这个其实类似于莫队的区间维护

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
int t, n, m, ans[maxn];
struct node {
    int l, r, flag;
} qu[maxn];
int cmp(node a, node b) {
    if (a.l == b.l) return a.r < b.r;
    return a.l < b.l;
}
int main() {
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &m);
        for (int i = 0 ; i < m ; i++) {
            scanf("%d%d", &qu[i].l, &qu[i].r);
            qu[i].flag = 0;
        }
        sort(qu, qu + m, cmp);
        set<int>st;
        for (int i = 1 ; i <= n ; i++) {
            st.insert(i);
            ans[i] = 1;
        }
        for (int i = qu[0].l ; i <= qu[0].r ; i++) {
            ans[i] = *st.begin();
            st.erase(st.begin());
        }
        int L = qu[0].l, R = qu[0].r;
        for (int i = 1 ; i < m ; i++) {
            while(L < qu[i].l) {
                st.insert(ans[L]);
                L++;
            }
            while(R < qu[i].r) {
                if (R < qu[i].l-1) R++;
                else {
                    R++;
                    ans[R] = *st.begin();
                    st.erase(st.begin());
                }
            }
        }
        for (int i = 1 ; i <= n ; i++) {
            if (i != n)  printf("%d ", ans[i]);
            else printf("%d\n", ans[i]);
        }
    }
    return 0;
}