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 jholds.
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 6.

 

 

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．

 

 1 #include <bits/stdc++.h>
 2 #define N 100005
 3 using namespace std;
 4 struct Node{
 5     int x,y;
 6 
 7 }node[N];
 8 bool cmp(Node a,Node b){
 9     if(a.x==b.x){
10         return a.y>b.y;
11     }
12     return a.x<b.x;
13 }
14 set<int> s;
15 int ans[N];
16 int n;
17 int main(){
18     scanf("%d",&n);
19     while(n--){
20         s.clear();
21         int m,k;
22         scanf("%d%d",&m,&k);
23         for(int i=1;i<=m;++i)
24             s.insert(i);
25         for(int i=0;i<k;++i){
26             scanf("%d%d",&node[i].x,&node[i].y);
27         }
28         sort(node,node+k,cmp);
29         int xn = 1,yn = 0;
30         for(int i=0;i<k;++i){
31             if(node[i].y<=yn)
32                 continue;
33             if(yn>=node[i].x){
34                 while(xn<node[i].x)
35                     s.insert(ans[xn++]);
36                 while (yn<node[i].y) {
37                     ans[++yn] = *s.begin();
38                     s.erase(s.begin());
39                 }
40             }else if(yn<node[i].x){
41                 while(yn<node[i].x)
42                     ans[++yn] = 1;
43                 while(xn<node[i].x)
44                     s.insert(ans[xn++]);
45                 if((*s.begin())==1)
46                     s.erase(s.begin());
47                 while(yn<node[i].y){
48                     ans[++yn] = (*s.begin());
49                     s.erase(s.begin());
50                 }
51             }
52         }
53         while(yn<m)
54             ans[++yn] = 1;
55         for(int i=1;i<=m;i++)
56             printf("%d%c", ans[i], i==m?'\n':' ');
57     }
58     return 0;
59 }

 

 

 

下面是我不过的代码：

 1 #include <bits/stdc++.h>
 2 #define N 200005
 3 using namespace std;
 4 struct Node{
 5     int x,y;
 6 
 7 }node[N];
 8 bool cmp(Node a,Node b){
 9     if(a.x==b.x){
10         return a.y>b.y;
11     }
12     return a.x<b.x;
13 }
14 set<int> s;
15 int ans[N];
16 int n;
17 int main(){
18     scanf("%d",&n);
19     while(n--){
20         for(int i=1;i<100005;i++)
21             s.insert(i);
22         int m,k;
23         scanf("%d%d",&m,&k);
24         for(int i=0;i<k;i++){
25             scanf("%d%d",&node[i].x,&node[i].y);
26         }
27         sort(node,node+k,cmp);
28         for(int j = node[0].x;j<=node[0].y;j++){
29             ans[j] = *s.begin();
30             s.erase(s.begin());
31         }
32         int xn = node[0].x,yn = node[0].y;
33         for(int i=1;i<k;i++){
34             if(node[i].y<=yn)
35                 continue;
36             if(yn>=node[i].x&&node[i].y>yn){
37                 for(int j = xn;j<node[i].x;j++){
38                     s.insert(ans[j]);
39                 }
40                 for(int j = yn+1;j<=node[i].y;j++){
41                     ans[j] = *s.begin();
42                     s.erase(s.begin());
43                 }
44                 xn = node[i].x,yn = node[i].y;
45             }else if(yn<node[i].x){
46                 for(int j = xn;j<=yn;j++){
47                     s.insert(ans[j]);
48                 }
49                 for(int j = node[i].x;j<=node[i].y;j++){
50                     ans[j] = *s.begin();
51                     s.erase(s.begin());
52                 }
53                 xn = node[i].x,yn = node[i].y;
54             }
55         }
56         for(int i=1;i<=m;i++){
57             if(ans[i]==0){
58                 printf("1 ");
59             }else{
60                 printf("%d ",ans[i]);
61             }
62         }
63         printf("\n");
64         s.clear();
65     }
66     return 0;
67 }