The kth great number_优先队列_priority

The kth great number

TimeLimit: 2000/1000 MS (Java/Others)  MemoryLimit: 65768/65768 K (Java/Others)

64-bit integer IO format:%I64d

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. 

SampleInput

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

SampleOutput

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).

题意大意， 就是给 一段数列， 求固定的第几大数， 细节请看 sampleinput
这题是优先队列的模板题。 当然是可以用 C ++ 的 stl。 但还是想了想， 自己搞下吧。
code：

#include <cstdio>
#include <algorithm>
#include <ctime>
#include <cstdlib>
using namespace std;
typedef int elemType;
const elemType all = 1000005;
elemType M1[ all ];
// 运用函数指针 和 类型重定义， 可以封装成 函数模板。
void heap_push( elemType [], int, bool (*cmp)( elemType, elemType ) );
void heap_pop( elemType [], int, bool (*cmp)( elemType, elemType ) );
bool cmp_min( elemType elema, elemType elemb );

int main(void)
{
    int m, n, tmp;
    char ch[5];

    while( scanf( "%d%d", &n, &m ) != EOF ){
        for( int i=0; i < m; ++ i ){
            scanf( "%s%d", ch, M1+i );
            heap_push( M1, i, cmp_min );
        }

        for( int i=m; i < n; ++ i ){
            scanf( "%s", ch );
            if( ch[0] == 'Q' ){
                printf( "%d\n", M1[0] );
            }
            else{
                scanf( "%d", &tmp );
                if( cmp_min( M1[0], tmp ) ){
                    heap_pop( M1, m-1, cmp_min );
                    M1[ m-1 ] = tmp;
                    heap_push( M1, m-1, cmp_min );
                }
            }
        }
    }

    return 0;
}

bool cmp_min( elemType elema, elemType elemb )
{
    return elema < elemb;
}

void heap_push( elemType container[], int num, bool (*cmp)( elemType elema, elemType elemb ) )
{
    while( num && cmp( container[ num ], container[ (num-1)/2 ] ) ){
        swap( container[ num ], container[ (num-1)/2 ] );
        num = (num-1)/2;
    }
}

void heap_pop( elemType container[], int num, bool (*cmp)( elemType elema, elemType elemb ) )
{
    swap( container[0], container[num] );
    -- num;
    elemType spot = 0;
    bool isreturn = false;
    while( 2*spot+2 <= num && ( cmp( container[2*spot+1], container[spot] ) || cmp( container[ 2*spot+2 ], container[spot] ) ) ){
        if( cmp( container[ 2*spot+1 ], container[ 2*spot+2 ] ) ){
            swap( container[ 2*spot+1 ], container[ spot ] );
            spot = 2*spot+1;
        }

        else{
            swap( container[ 2*spot+2 ], container[ spot ] );
            spot = 2*spot+2;
        }
        isreturn = true;
    }

    if( ! isreturn && 2*spot+1 <= num && cmp( container[ 2*spot+1 ], container[ spot ] ) ){
        swap( container[ spot ], container[ 2*spot+1 ] );
    }
}