| 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. 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: 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 Recommend liuyiding |
// D
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
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;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
const int N=101000;
int _,n,m,pre[N],l,r,ret[N];
int main() {
/*分析样例5 2
1 3
2 4
*/
for (scanf("%d",&_);_;_--) {
scanf("%d%d",&n,&m);
rep(i,1,n+1) pre[i]=i;//pre : 1 2 3 4 5
rep(i,0,m) {
scanf("%d%d",&l,&r);
pre[r]=min(pre[r],l);//pre :1 2 1 2 5初步得到对应位置最开始影响的坐标位置
}
per(i,1,n) pre[i]=min(pre[i],pre[i+1]); //pre :1 1 1 2 5 得到最终的影响区间
int pl=1;
set<int> val;
rep(i,1,n+1) val.insert(i); // val : 1 2 3 4 5
rep(i,1,n+1) { // 到4坐标之前都是在同一个区间所以前面是1 2 3
while (pl<pre[i]) { // 当它到达4位置时可以知道它最早的影响区间为2下标，
val.insert(ret[pl]);// 所以它所在位置在小于ret[2]都可以赋值（也就是在val中加入x，x<ret[2]
// 所以ret为1，pl为2
pl++; // 当到5下标同理可加入pl到pre[5]之前的
} // 其实比如5时输入区间没有包含的所以它一定时1，而
ret[i]=*val.begin(); // 从pl到改位置的ret又使得val又变成了1~n
val.erase(ret[i]); // 为什么呢？因为到5时（也就是到输入没有包含的区间时，
// 发生了断层，所以所以后面的都是从1开始赋值
}
rep(i,1,n+1) printf("%d%c",ret[i]," \n"[i==n]);
}
}