HDU6301 Distinct Values （多校第一场1004） （贪心）

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 <bits/stdc++.h>
 3 using namespace std;
 4 #define rep(i,a,n) for (int i=a;i<n;i++)
 5 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 6 #define pb push_back
 7 #define mp make_pair
 8 #define all(x) (x).begin(),(x).end()
 9 #define fi first
10 #define se second
11 #define sz(x) ((int)(x).size())
12 typedef vector<int> vi;
13 typedef long long ll;
14 typedef pair<int,int> 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 // head
19 
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<int> 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<pre[i]) {//小于pre[i]的点的值  插入set
38                 val.insert(ret[pl]);
39                 pl++;
40             }
41             ret[i]=*val.begin();//不小于直接取最小的数放进去
42             val.erase(ret[i]);//删除刚放入的数
43         }
44         rep(i,1,n+1) printf("%d%c",ret[i]," \n"[i==n]);
45     }
46 }