# 【最大访客数】

/*

（由小到大），道理很简单，只要先计算某时之前总共来访了多少访客，然后再减去某时之前的离开访客，就可以轻易的解出这个问题。
*/

#include <stdio.h>
#include <stdlib.h>

#define MAX 100
#define SWAP(x, y) {int t; t = x; x = y; y = t;}

int partition(int [], int , int );
void quicksort(int [], int , int );
int maxguest(int [], int[], int, int );

int main(void)
{
int x[MAX] = {0};
int y[MAX] = {0};
int time = 0;
int count = 0;

printf("\n 输入来访与离开125;时间(0~24): ");
printf("\n范例: 10 15");
printf("\n输入-1 -1结束\n");
while(count < MAX)
{
printf("\n>>");
scanf("%d%d", &x[count], &y[count]);
if(x[count] < 0)
{
break;
}
count++;
}

if(count >= MAX)
{
printf("\n超出最大访客数(%d)", MAX);
count--;
}

quicksort(x, 0, count);
quicksort(y, 0, count);

while(time < 25)
{
printf("\n %d 时的最大访客数 : %d", time, maxguest(x, y, count, time));
time++;
}
printf("\n");
return 0;
}

int maxguest(int x[], int y[], int count, int time)
{
int i, num = 0;
for(i = 0; i <= count; i++)
{
if(time > x[i])
{
num++;
}
if(time > y[i])
{
num--;
}
}
return num;
}

int partition(int number[], int left, int right)
{
int i, j, s;
s = number[right];
i = left - 1;
for(j = left; j < right; j++)
{
if(number[j] <= s)
{
i++;
SWAP(number[i], number[j]);
}
}
SWAP(number[i+1], number[right]);
return i+1;
}

void quicksort(int number[], int left, int right)
{
int q;
if(left < right)
{
q = partition(number, left, right);
quicksort(number, left, q - 1);
quicksort(number, q + 1, right);
}
}

posted @ 2017-02-01 17:10  天秤libra  阅读(1163)  评论(0编辑  收藏