HDU 4883 TIANKENG’s restaurant （区间更新）

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

这道题目很容易想到用线段树更新，比赛的时候也确实用线段树过了，但实际上用线段树更新有些大材小用，因为题目只需查询一次，并没有动态更新和动态查询。

可以用一个sum数组记录该点总共的人数，输入的时候可预处理第i批客人的变化情况。来的时候  sum[i]就加上相应的人数，走的时候sum[i]就减去相应的人数。

可以得到地推公式 sum[i] += sum[i-1]; i表示时间的推移，从第0分钟一直可以推到结束时间。因为期间不用再次更新也无需查询，这种算法要比线段树更具优势。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stack>
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;
const int maxn = 1500;
const int MAX = 0x3f3f3f3f;
const int mod = 1000000007;
int t, n;
int sum[maxn];
int main()
{
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        memset(sum, 0, sizeof(sum));
        for(int i = 0; i < n; i++) {
            int tmp, h1, m1, h2, m2;
            scanf("%d %d:%d %d:%d", &tmp, &h1, &m1, &h2, &m2);
            int st = h1*60+m1;
            int en = h2*60+m2;
            sum[st] += tmp;
            sum[en] -= tmp;
        }
        int ans = sum[0];
        for(int i = 1; i <= 1440; i++) {
            sum[i] += sum[i-1];
            ans = max(ans, sum[i]);
        }
        printf("%d\n", ans);
    }
    return 0;
}



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