Distinct Values

Distinct Values

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2858    Accepted Submission(s): 923

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≠ajholds.
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

 

 

Source

2018 Multi-University Training Contest 1

 

 

这题看起来用数组模拟都可以，但是会超时．

不过做了不知道为什么不过．

 

#include <bits/stdc++.h>
#define N 100005
using namespace std;
struct Node{
    int x,y;

}node[N];
bool cmp(Node a,Node b){
    if(a.x==b.x){
        return a.y>b.y;
    }
    return a.x<b.x;
}
set<int> s;
int ans[N];
int n;
int main(){
    scanf("%d",&n);
    while(n--){
        s.clear();
        int m,k;
        scanf("%d%d",&m,&k);
        for(int i=1;i<=m;++i)
            s.insert(i);
        for(int i=0;i<k;++i){
            scanf("%d%d",&node[i].x,&node[i].y);
        }
        sort(node,node+k,cmp);
        int xn = 1,yn = 0;
        for(int i=0;i<k;++i){
            if(node[i].y<=yn)
                continue;
            if(yn>=node[i].x){
                while(xn<node[i].x)
                    s.insert(ans[xn++]);
                while (yn<node[i].y) {
                    ans[++yn] = *s.begin();
                    s.erase(s.begin());
                }
            }else if(yn<node[i].x){
                while(yn<node[i].x)
                    ans[++yn] = 1;
                while(xn<node[i].x)
                    s.insert(ans[xn++]);
                if((*s.begin())==1)
                    s.erase(s.begin());
                while(yn<node[i].y){
                    ans[++yn] = (*s.begin());
                    s.erase(s.begin());
                }
            }
        }
        while(yn<m)
            ans[++yn] = 1;
        for(int i=1;i<=m;i++)
            printf("%d%c", ans[i], i==m?'
':' ');
    }
    return 0;
}

下面是我不过的代码：

#include <bits/stdc++.h>
#define N 200005
using namespace std;
struct Node{
    int x,y;

}node[N];
bool cmp(Node a,Node b){
    if(a.x==b.x){
        return a.y>b.y;
    }
    return a.x<b.x;
}
set<int> s;
int ans[N];
int n;
int main(){
    scanf("%d",&n);
    while(n--){
        for(int i=1;i<100005;i++)
            s.insert(i);
        int m,k;
        scanf("%d%d",&m,&k);
        for(int i=0;i<k;i++){
            scanf("%d%d",&node[i].x,&node[i].y);
        }
        sort(node,node+k,cmp);
        for(int j = node[0].x;j<=node[0].y;j++){
            ans[j] = *s.begin();
            s.erase(s.begin());
        }
        int xn = node[0].x,yn = node[0].y;
        for(int i=1;i<k;i++){
            if(node[i].y<=yn)
                continue;
            if(yn>=node[i].x&&node[i].y>yn){
                for(int j = xn;j<node[i].x;j++){
                    s.insert(ans[j]);
                }
                for(int j = yn+1;j<=node[i].y;j++){
                    ans[j] = *s.begin();
                    s.erase(s.begin());
                }
                xn = node[i].x,yn = node[i].y;
            }else if(yn<node[i].x){
                for(int j = xn;j<=yn;j++){
                    s.insert(ans[j]);
                }
                for(int j = node[i].x;j<=node[i].y;j++){
                    ans[j] = *s.begin();
                    s.erase(s.begin());
                }
                xn = node[i].x,yn = node[i].y;
            }
        }
        for(int i=1;i<=m;i++){
            if(ans[i]==0){
                printf("1 ");
            }else{
                printf("%d ",ans[i]);
            }
        }
        printf("
");
        s.clear();
    }
    return 0;
}