2015 Multi-University Training Contest 5 hdu 5349 MZL's simple problem

MZL's simple problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 740    Accepted Submission(s): 357

Problem Description

A simple problem
Problem Description
You have a multiple set,and now there are three kinds of operations:
1 x : add number x to set
2 : delete the minimum number (if the set is empty now,then ignore it)
3 : query the maximum number (if the set is empty now,the answer is 0)

 

Input

The first line contains a number N (N≤106),representing the number of operations.
Next N line ,each line contains one or two numbers,describe one operation.
The number in this set is not greater than 109.

 

Output

For each operation 3,output a line representing the answer.

 

Sample Input

6

1 2

1 3

3

1 3

1 4

3

 

Sample Output

3 4

 

Source

2015 Multi-University Training Contest 5

 

解题：由于我们只需要查询最大值，所以最大值总是最后一个被删的，如果当前集合只有一个元素，那么肯定是删最大的元素了，否则，随便删就是了，不影响最大值

 

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n,op,val;
    while(~scanf("%d",&n)) {
        int sz = 0,maxval = INT_MIN;
        while(n--) {
            scanf("%d",&op);
            switch(op) {
            case 1:
                scanf("%d",&val);
                maxval = max(val,maxval);
                sz++;
                break;
            case 2:
                sz = max(0,sz-1);
                if(!sz) maxval = INT_MIN;
                break;
            case 3:
                printf("%d
",sz?maxval:0);
                break;
            default:
                ;
            }
        }
    }
    return 0;
}