Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

If you think participating in a programming contest is stressful, imagine being an air traffic controller. With human lives at stake, an air traffic controller has to focus on tasks while working under constantly changing conditions as well as dealing with unforeseen events.

Consider the task of scheduling the airplanes that are landing at an airport. Incoming airplanes report their positions, directions, and speeds, and then the controller has to devise a landing schedule that brings all airplanes safely to the ground. Generally, the more time there is between successive landings, the ``safer" a landing schedule is. This extra time gives pilots the opportunity to react to changing weather and other surprises.

Luckily, part of this scheduling task can be automated - this is where you come in. You will be given scenarios of airplane landings. Each airplane has a time window during which it can safely land. You must compute an order for landing all airplanes that respects these time windows. Furthermore, the airplane landings should be stretched out as much as possible so that the minimum time gap between successive landings is as large as possible. For example, if three airplanes land at 10:00am, 10:05am, and 10:15am, then the smallest gap is five minutes, which occurs between the first two airplanes. Not all gaps have to be the same, but the smallest gap should be as large as possible.

Input 

The input file contains several test cases consisting of descriptions of landing scenarios. Each test case starts with a line containing a single integer n (2n8), which is the number of airplanes in the scenario. This is followed by n lines, each containing two integers a_i, b_i, which give the beginning and end of the closed interval [a_i, b_i] during which the i-th plane can land safely. The numbers a_i and b_i are specified in minutes and satisfy 0a_ib_i1440.

The input is terminated with a line containing the single integer zero.

Output 

For each test case in the input, print its case number (starting with 1) followed by the minimum achievable time gap between successive landings. Print the time split into minutes and seconds, rounded to the closest second. Follow the format of the sample output.

Sample Input 

3 
0 10 
5 15 
10 15 
2 
0 10 
10 20 
0

Sample Output 

Case 1: 7:30 
Case 2: 20:00

---

題目描述：

現在有 n (<= 8)台飛機在機場上盤旋，每台飛機著陸時間為一個區間 [a, b] (單位：分鐘)

機場人員希望它們著陸時間間隔最小值最大化。

四捨五入到秒，輸出格式  分：秒

題目解法：

窮舉每台飛機著陸的順序，二分間隔的時間，使用 Greedy 檢查。

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
int n, a[10], b[10], p[10];
int check(double gap) {
    double now = a[p[0]]+gap;
    int i;
    for(i = 1; i < n; i++) {
        if(b[p[i]] < now)
            return 0;
        if(a[p[i]] > now)
            now = a[p[i]];
        now += gap;
    }
    return 1;
}
double sol() {
    double l, r, mid;
    l = 0, r = 1440*60;//seconds
    // binary search max gap
    while(fabs(l-r) > 1e-2) {
        mid = (l+r)/2;
        if(check(mid))
            l = mid;
        else
            r = mid;
    }
    return mid;
}
int main() {
    int i, cases = 0;
    while(scanf("%d", &n) == 1 && n) {
        for(i = 0; i < n; i++) {
            scanf("%d %d", &a[i], &b[i]);
            a[i] *= 60, b[i] *= 60;
        }
        for(i = 0; i < n; i++)
            p[i] = i;
        double ret = 0;
        do {
            double tmp = sol();
            ret = max(ret, tmp);
        } while(next_permutation(p, p+n));
        printf("Case %d: ", ++cases);
        int seconds = (int)round(ret);
        printf("%d:%02d\n", seconds/60, seconds%60);
    }
    return 0;
}