简单模拟,题意描述不清晰,我来仔细描述一下,flashget在下载一些文件,每个文件有一个初始大小,初始速度,最大速度。当一个文件下载结束后,他的速度(占用的带宽)会平均分配给别的任务(各文件的速度增加值相同),假设我们要把速度x分配给a个文件,那么就要给每个文件分配x/a的速度,前提条件是分配后该文件的速度不超过其最大速度。若超过了则不能给他分配这么多,只让它达到其最大速度即可。这样一来我们的下载速度就会有剩余,因为受到最大速度的限制,有些文件并没有被分配到x/a这么多的速度。对于剩余的速度怎么处理呢?就是除去那几个已经到达最大速度的文件,其余文件按照刚才的方法再分配一次剩余速度。模拟即可。这题精度很难控制,我开始试着用先将所有到达最大速度的进程先处理,最后再计算平均值的方法,居然是错的,理论上这种做法比上述做法精度高。
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#include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define maxn 105 struct Elem { double size, speed, max_speed; } elem[maxn]; int n, m; double time_left[maxn]; bool finished[maxn]; double ans[maxn]; double reached[maxn]; int task_left; void input() { scanf("%d", &m); for (int i = 0; i < n; i++) scanf("%lf%lf%lf", &elem[i].size, &elem[i].speed, &elem[i].max_speed); } void cal_time() { for (int i = 0; i < n; i++) if (!finished[i]) time_left[i] = elem[i].size / elem[i].speed; } void size_reduce(double min_time) { for (int i = 0; i < n; i++) if (!finished[i]) elem[i].size -= min_time * elem[i].speed; } int get_min() { int ret = 0; while (finished[ret]) ret++; for (int i = ret + 1; i < n; i++) if (!finished[i] && time_left[i] < time_left[ret]) ret = i; return ret; } void distribute(double bandwidth) { double band_left = bandwidth; int num = 0; for (int i = 0; i < n; i++) if (!finished[i] && !reached[i]) num++; if (num == 0) return; double average; while (band_left > 0) { average = band_left / num; bool did = false; for (int i = 0; i < n; i++) { if (finished[i] || reached[i]) continue; did = true; if (elem[i].max_speed - elem[i].speed < average) { band_left -= elem[i].max_speed - elem[i].speed; elem[i].speed = elem[i].max_speed; reached[i] = true; num--; }else { band_left -= average; elem[i].speed += average; } } if (!did) break; } } void work() { memset(finished, 0, sizeof(finished)); memset(reached, 0, sizeof(reached)); double now_time = 0; task_left = n; while (task_left > 0) { cal_time(); int next_task = get_min(); finished[next_task] = true; task_left--; double bandwidth = elem[next_task].speed; double min_time = time_left[next_task]; size_reduce(min_time); now_time += min_time; ans[next_task] = now_time; distribute(bandwidth); } } int main() { //freopen("t.txt", "r", stdin); int t = 0; while (scanf("%d", &n), n) { t++; printf("Case %d:\n", t); input(); work(); for (int i = 0; i < n; i++) printf("NO%d:%.3fs\n", i + 1, ans[i]); } return 0; }