HDU 多校对抗赛 D Distinct Values

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1311 Accepted Submission(s): 398

Problem Description

Chiaki has an array of

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

Output

For each test case, output

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

题意：你需要构造一个长度为n的字典序最小的序列，要求如下

　　　1，有m个区间，在这些区间内，数字不得重复

　　　2，长度为n，不能有0出现

题解：既然要求字典序最小，那么我们那些没有被区间覆盖的地方一定是1

　　　区间覆盖的地方得特殊考虑，区间是否相交（重叠）等问题

　　　于是我们先把区间按照左端点排序，每次在区间内放尽可能小的就行

　　　格式要求行尾不能有多余的空格

代码如下

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn = 1e5+5;
struct node{
  int l,r;
}a[maxn];
bool cmp(node a,node b){
  if(a.l==b.l) return a.r<b.r;
  return a.l<b.l;
}
int ans[maxn];
int vis[maxn];
int main(){
  int T;
  scanf("%d",&T);
  while(T--){
    int n,m;
    memset(vis,0,sizeof(vis));
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
      scanf("%d%d",&a[i].l,&a[i].r);
    }
    sort(a+1,a+m+1,cmp);
    priority_queue<int,vector<int>,greater<int> >q;
        for(int i=1; i<=n; i++){
            ans[i]=1;vis[i]=0;
            q.push(i);
        }
        int l=a[1].l,r=a[1].l;
        for(int i=1;i<=m;i++){
            for(;l<a[i].l;l++){
                if(vis[l]) q.push(ans[l]);
            }
            for(;r<=a[i].r;r++){
                if(r>=a[i].l){
                    ans[r]=q.top();q.pop();
                    vis[r]=1;
                }
            }
        }
        for(int i=1; i<=n; i++) {
            printf("%d",ans[i]);
            if(i==n) printf("\n");
            else printf(" ");
        }
  }
}

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