hdu4006 The kth great number 优先队列

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The kth great number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 10328 Accepted Submission(s): 4138

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

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

两个闲人在玩游戏，每回合A可以写数字或者询问B第k大的数字是啥

我们要做的就是回答第k大的数字是啥

之前各种写都超时啊。。。机房的空调好冷啊。。。。。

这题好早写的我都忘了是啥了。。。。java编译器怎么还没装好啊。。。

反正之前做过这样的题啊，就是sliding window 啊。。。。那玩意儿好像还更难写来着。。。

优先队列里数越小优先值越高要在定义的时候加个greater<int> 啊

具体看代码：

/*
 ━━━━━┒
 ┓┏┓┏┓┃μ'sic foever!!
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
using namespace std;
const int maxn=1000010;
const int inf=1<<30;
int c[maxn];
int lowbit(int x){
    return x&(-x);
}

void update(int x,int val){
    while(x<=maxn){
        c[x] += val;
        x += lowbit(x);
    }
}


int getsum(int x){
    int sum=0;
    while(x>0){
        sum += c[x];
        x -=lowbit(x);
    }
    return sum;
}
bool compare(int a,int b){
    return a>b;
}

int main(){
    int k,n,i,j,number;
    string str;
while(~scanf("%d%d",&n,&k)){
    priority_queue<int,vector<int>,greater<int> > save;
    for(i=1;i<=n;i++){
        cin>>str;
        if(str=="I"){
            scanf("%d",&number);
            if(save.size()<k){
                save.push(number);
            }else{
                if(number>save.top()){
                    save.pop();
                    save.push(number);
                }
            }
        }else{
            printf("%d\n",save.top());
        }
    }
}

    return 0;
}