sdutoj 3252 Lowest Unique Price
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题目链接:点击打开链接
题目大意:
Recently my buddies and I came across an idea! We want to build a website to sell things in a new way.
For each product, everyone could bid at a price, or cancel his previous bid, finally we sale the product to the one who offered the "lowest unique price". The lowest unique price is defined to be the lowest price that was called only once.
So we need a program to find the "lowest unique price", We'd like to write a program to process the customers' bids and answer the query of what's the current lowest unique price.
All what we need now is merely a programmer. We will give you an "Accepted" as long as you help us to write the program.
输入
The first line of input contains an integer T, indicating the number of test cases (T ≤ 60).
Each test case begins with a integer N (1 ≤ N ≤ 200000) indicating the number of operations.
Next N lines each represents an operation.
There are three kinds of operations:
"b x": x (1 ≤ x ≤ 106) is an integer, this means a customer bids at price x.
"c x": a customer has canceled his bid at price x.
"q" : means "Query". You should print the current lowest unique price.
Our customers are honest, they won\'t cancel the price they didn't bid at.
输出
Please print the current lowest unique price for every query ("q"). Print "none" (without quotes) if there is no lowest unique price.
示例输入
2 3 b 2 b 2 q 12 b 2 b 2 b 3 b 3 q b 4 q c 4 c 3 q c 2 q
示例输出
none none 4 3 2
思路:用哈希存储,遍历查找
#include <iostream> 02.#include<cstdio> 03.#include<cstdlib> 04.#include<cstring> 05. 06.using namespace std; 07.int a[1000005]; 08.int main() 09.{ 10. int t; 11. int n; 12. 13. scanf("%d",&t); 14. char s[2]; 15. int x; 16. int mx=-1; 17. while(t--) 18. { 19. mx=-1; 20. memset(a,0,sizeof(a)); 21. scanf("%d",&n); 22. for(int i=0;i<n;i++) 23. { 24. scanf("%s",s); 25. if(s[0]=='b') 26. { 27. scanf("%d",&x); 28. if(x>mx)mx=x; 29. a[x]++; 30. } 31. else if(s[0]=='c') 32. { 33. scanf("%d",&x); 34. a[x]--; 35. } 36. else if(s[0]=='q') 37. { 38. int f=0; 39. int j; 40. for(j=0;j<=mx;j++) 41. { 42. if(a[j]==1) 43. { 44. f=1; 45. break; 46. } 47. } 48. if(f==1)printf("%d\n",j); 49. else printf("none\n"); 50. } 51. } 52. } 53. return 0; 54.}
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