HDU4883TIANKENG’s restaurant

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于 2017-05-19 21:26:00 发布

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分类专栏：算法 ------贪心

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本文介绍了一个基于顾客到店和离店时间的算法，用于计算餐厅高峰期所需的最小椅子数量。通过贪心策略，按顾客到达时间排序，并使用优先级队列记录顾客离开时间，实现了资源的有效分配。

TIANKENG’s restaurant

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

Problem Description

TIANKENG manages a restaurant after graduating from ZCMU, and tens of thousands of customers come to have meal because of its delicious dishes. Today n groups of customers come to enjoy their meal, and there are Xi persons in the ith group in sum. Assuming that each customer can own only one chair. Now we know the arriving time STi and departure time EDi of each group. Could you help TIANKENG calculate the minimum chairs he needs to prepare so that every customer can take a seat when arriving the restaurant?

Input

The first line contains a positive integer T(T<=100), standing for T test cases in all.

Each cases has a positive integer n(1<=n<=10000), which means n groups of customer. Then following n lines, each line there is a positive integer Xi(1<=Xi<=100), referring to the sum of the number of the ith group people, and the arriving time STi and departure time Edi(the time format is hh:mm, 0<=hh<24, 0<=mm<60), Given that the arriving time must be earlier than the departure time.

Pay attention that when a group of people arrive at the restaurant as soon as a group of people leaves from the restaurant, then the arriving group can be arranged to take their seats if the seats are enough.

Output

For each test case, output the minimum number of chair that TIANKENG needs to prepare.

Sample Input

2
2
6 08:00 09:00
5 08:59 09:59
2
6 08:00 09:00
5 09:00 10:00

Sample Output

11
6

题意：有个饭店，每一组客人来的时间和离开的时间不同，每组客人的人数可能不同。需要计算准备的椅子数目。

题解：贪心做法：按来的时间将客人排序，遍历数组。把来的人加入优先级队列，离开时间早的在队头。

遍历数组时，比较当前客人到的时间是否大于等于队列顶端客人离开的时间，如果是，则出队表示客人离开。

遍历时用个sum来记录当前座位总数，用Max记录到目前位置需要的最大座位数目。

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#define N 10005
using namespace std;
struct time{
	int s, e, cnt;
};
bool operator < (const time &a, const time &b){
	return a.e > b.e;
}
bool cmp(const time &a, const time &b){
	return a.s < b.s;
}
time T[N];
priority_queue<time> Q;
int main(){
	int t, n;
	int a, b;
	time p;
	cin>>t;
	while(t--){
		scanf("%d", &n);
		for(int i = 0; i < n; i++){
			scanf("%d", &T[i].cnt);
			scanf("%d:%d", &a, &b);
			T[i].s = a*60 + b;
			scanf("%d:%d", &a, &b);
			T[i].e = a*60 + b;
		}
		sort(T, T+n, cmp);
		int Max = 0, sum = 0;
		for(int i = 0; i <n; i++){
			while(!Q.empty() && Q.top().e <= T[i].s){
				p = Q.top();
				Q.pop();
				sum -= p.cnt;
			}
			Q.push(T[i]);
			sum += T[i].cnt;
			Max = max(Max, sum);
		}
		printf("%d\n", Max);
	}
	return 0;
}