20140916 【 贪心 】 bestcoder #2 + hdoj 4883 TIANKENG’s restaurant

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TIANKENG’s restaurant

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 829    Accepted Submission(s): 375

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

226 08:00 09:005 08:59 09:5926 08:00 09:005 09:00 10:00

 

Sample Output

116

 

Source

BestCoder Round #2

 

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这种题目，采用贪心的方法。

起点 + n

终点 - n

然后对所有时间点扫一遍（进行累加）

过程中得到的最大值就是正解。。。

（注意：起始与终止时间点重合时，要怎么处理）

在本题中：

起止重合时，

允许先减，不允许先加。

所以：

1. 要么对时间点排序时，时间重叠的，先排减时间。

2. 要么累加时间点的值时，把同一时间点的值先累加起来。

如下两份代码。。。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define MAXN 10100#define MIN 0struct PP{    int n, t;}a[MAXN*2];bool cmp(const PP a, const PP b){    if( a.t==b.t )            return a.n<b.n;    return a.t<b.t;}int n;int main(){        int T;    while( EOF != scanf("%d\n", &T) ){        while( T-- ){            scanf("%d\n", &n);            int m = n<<1;            for(int i=0, s,e; i<n; i+=1){                scanf("%d %2d:%2d %2d:%2d\n", &a[i].n, &a[i].t, &s, &a[i+n].t, &e);                a[i].t = a[i].t*60 + s;                a[i+n].n = -a[i].n;                a[i+n].t = a[i+n].t*60 + e;            }            sort(a, a+m, cmp);            long long ans=MIN, now=MIN;            for(int i=0; i<m; i++){                now += a[i].n;                ans = max(ans, now);            }            printf("%lld\n", ans);        }    }    return 0;}

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define MAXN 10100#define MIN 0struct PP{    int n, t;}a[MAXN*2];bool cmp(const PP a, const PP b){   return a.t<b.t; }int n;int main(){         int T;    while( EOF != scanf("%d\n", &T) ){        while( T-- ){            scanf("%d\n", &n);            for(int i=0, s,e; i<n; i++){                scanf("%d %2d:%2d %2d:%2d\n", &a[i].n, &a[i].t, &s, &a[i+n].t, &e);                a[i].t = a[i].t*60 + s;                a[i+n].n = -a[i].n;                a[i+n].t = a[i+n].t*60 + e;            }            int m = n<<1;            sort(a, a+m, cmp);            long long ans=MIN, now=MIN;            for(int i=0, k; i<m; i=k){                k = i+1;                while( k<m && a[i].t==a[k].t ){                    a[i].n += a[k].n;                    k++;                }                now += a[i].n;                ans = max(ans, now);            }            printf("%lld\n", ans);        }    }    return 0;}