Distinct Values

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4869    Accepted Submission(s): 1659

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

 

 

给出q个[l,r]   要求[l,r]范围内数字不重复 ，求字典序最小的满足q个要求的序列。

 

 

首先预处理出来覆盖每一个点的最左端点pre[i] ，用p从1开始 和  pre[i]比较，(因为当前区间是[pre[i], i],  上个区间是 [p, i-1] )  如果 p < pre[i] , 就把[p, pre[i]-1] 的数都释放了（插入set还能用）如果 p > pre[i], 直接取set.begin().

 

 

 

 

1 // D 2 #include 
      
        3 using namespace std; 4 #define rep(i,a,n) for (int i=a;i
       
        =a;i--) 6 #define pb push_back 7 #define mp make_pair 8 #define all(x) (x).begin(),(x).end() 9 #define fi first10 #define se second11 #define SZ(x) ((int)(x).size())12 typedef vector
        
          VI;13 typedef long long ll;14 typedef pair
         
           PII;15 const ll mod=1000000007;16 ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){
    if(b&1)res=res*a%mod;a=a*a%mod;}return res;}17 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}18 // head19 20 const int N=101000;21 int _,n,m,pre[N],l,r,ret[N];//pre维护覆盖i的最左端点 22 int main() {23     for (scanf("%d",&_);_;_--) {24         scanf("%d%d",&n,&m);25         rep(i,1,n+1) pre[i]=i;26         rep(i,0,m) {27             scanf("%d%d",&l,&r);28             pre[r]=min(pre[r],l);29         per(i,1,n) pre[i]=min(pre[i],pre[i+1]);//pre[i]是pre[i]和pre[i+1]的最小值30         int pl=1;//从1开始 和覆盖每个点的最左端点pre[i]比较31         set
          
            val;32         rep(i,1,n+1) val.insert(i);//维护最小可用的数33         rep(i,1,n+1) {34             //上个 [pl, i-1]35 36             //当前 [pre[i], i]37             while (pl