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 + ;
 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 =  ; i < m ; i++) {
             scanf("%d%d", &qu[i].l, &qu[i].r);
             qu[i].flag = ;
         }
         sort(qu, qu + m, cmp);
         set<int>st;
         for (int i =  ; i <= n ; i++) {
             st.insert(i);
             ans[i] = ;
         }
         for (int i = qu[].l ; i <= qu[].r ; i++) {
             ans[i] = *st.begin();
             st.erase(st.begin());
         }
         int L = qu[].l, R = qu[].r;
         for (int i =  ; i < m ; i++) {
             while(L < qu[i].l) {
                 st.insert(ans[L]);
                 L++;
             }
             while(R < qu[i].r) {
                 if (R < qu[i].l-) R++;
                 else {
                     R++;
                     ans[R] = *st.begin();
                     st.erase(st.begin());
                 }
             }
         }
         for (int i =  ; i <= n ; i++) {
             if (i != n)  printf("%d ", ans[i]);
             else printf("%d\n", ans[i]);
         }
     }
     return ;
 }