HDOJ4006 The kth great number 【串的更改和维护】

The kth great number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 6121    Accepted Submission(s): 2471

Problem Description

Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.

 

Input

There are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number. 

 

Output

The output consists of one integer representing the largest number of islands that all lie on one line. 

 

Sample Input

8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q

 

Sample Output

1
2
3

Hint

Xiao Ming won't ask Xiao Bao the kth great number when the number of the written number is smaller than k. (1=<k<=n<=1000000).

 

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

 

 

解题思路： 

　　我的想法就是，既然是要求第 k 大的数字，我就是设一数组总长度为k，其中最小的元素值即为第k大的数

　　不需要优先队列 或者是 线段树、红黑树等一些高级数据结构就能写出来的一题。

 

 

我的代码：

 #include<stdio.h>
 #include<string.h>
 #include<math.h>
 #include<algorithm>
 using namespace std;
 int b[];
 int main(){
     int i,j,t,n,k,num,flag_t,min,flag;
     char c;
     while(EOF != scanf("%d%d",&n,&k)){
         flag_t = ;
         min = ;
         while(n--){
             getchar();
             scanf("%c",&c);
             if(c == 'I'){
                 scanf("%d",&num);
                 if(flag_t < k){
                     b[flag_t] = num;
                     if(num < min){
                         min = num;
                         flag = flag_t;
                     }
                     flag_t++;
                 }
                 else if(num > min){
                     b[flag] = num;
                     min = ;
                     for(i=;i<k;i++){
                         if(b[i] < min){
                             min = b[i];
                             flag = i;
                         }
                     }
                 }
             }
             else if(c == 'Q'){
                 printf("%d\n",min);
             }
         }
     }
     return ;
 }