sicily 1046. Plane Spotting（排序求topN）

Description
Craig is fond of planes. Making photographs of planes forms a major part of his daily life. Since he tries to stimulate his social life, and since it’s quite a drive from his home to the airport, Craig tries to be very efficient by investigating what the optimal times are for his plane spotting. Together with some friends he has collected statistics of the number of passing planes in consecutive periods of fifteen minutes (which for obvious reasons we shall call ‘quarters’). In order to plan his trips as efficiently as possible, he is interested in the average number of planes over a certain time period. This way he will get the best return for the time invested. Furthermore, in order to plan his trips with his other activities, he wants to have a list of possible time periods to choose from. These time periods must be ordered such that the most preferable time period is at the top, followed by the next preferable time period, etc. etc. The following rules define which is the order between time periods:

1. A period has to consist of at least a certain number of quarters, since Craig will not drive three hours to be there for just one measly quarter.
2. A period P1 is better than another period P2 if:
* the number of planes per quarter in P1 is higher than in P2;
* the numbers are equal but P1 is a longer period (more quarters);
* the numbers are equal and they are equally long, but period P1 ends earlier.

Now Craig is not a clever programmer, so he needs someone who will write the good stuff: that means you. So, given input consisting of the number of planes per quarter and the requested number of periods, you will calculate the requested list of optimal periods. If not enough time periods exist which meet requirement 1, you should give only the allowed time periods.

Input
The input starts with a line containing the number of runs N. Next follows two lines for each run. The first line contains three numbers: the number of quarters (1–300), the number of requested best periods (1–100) and the minimum number of quarters Craig wants to spend spotting planes (1–300). The sec-nod line contains one number per quarter, describing for each quarter the observed number of planes. The airport can handle a maximum of 200 planes per quarter.

Output
The output contains the following results for every run:

* A line containing the text “Result for run <N>:” where <N> is the index of the run.

* One line for every requested period: “<F>-<L>” where <F> is first quarter and <L> is the last quarter of the period. The numbering of quarters starts at 1. The output must be ordered such that the most preferable period is at the top.

题意：按照他给的规则，求能看到最多飞机的时间区间。

思路就是枚举+排序+要top几输出top几

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#define eps 1e-6

struct period {
    int start, end;
    int len;
    double avg;  /* plane per quarter */
    bool operator<(const period& other) const {
        if (fabs(avg - other.avg) < eps) {  // this.ppq == other.ppq
            if (fabs(len - other.len) < eps) {
                return end < other.end;
            } else {
                return len > other.len;
            }
        } else {
            return avg > other.avg;
        }
    }
} periods[310 * 310];

int main(void) {
#ifdef JDEBUG
    freopen("1046.in", "r", stdin);
    freopen("1046.out", "w", stdout);
#endif
    int ppq[310];  // plane per quater
    int sum[310];  // planes in ppq[1~i]
    int t;
    scanf("%d", &t);

    for (int i = 1; i <= t; ++i) {
        sum[0] = 0;  // bound
        int total; // number of quarters
        int requested; // number of requested best periods
        int available; // minimum number of quarters spent on spotting planes

        scanf("%d %d %d", &total, &requested, &available);

        for (int j = 1; j <= total; ++j) {
            scanf("%d", &ppq[j]);
            sum[j] = sum[j-1] + ppq[j];
        }

        int counter = 0;

        // for every [start, end] in [1, total]
        // where start - end + 1 >= available
        for (int start = 1; start + available - 1 <= total; ++start) {
            for (int end = start + available - 1; end <= total; ++end) {
                periods[counter].start = start;
                periods[counter].end = end;
                periods[counter].len = end - start + 1;
                periods[counter].avg =
                    (double)(sum[end] - sum[start - 1]) / periods[counter].len;
                counter++;
            }
        }

        std::sort(periods, periods + counter);

        printf("Result for run %d:\n", i);
        for (int p = 0; p < requested && p < counter; ++p) {
            printf("%d-%d\n", periods[p].start, periods[p].end);
        }
    }

    return 0;
}