UVA - 10026 Shoemaker's Problem
2014-09-27 16:32
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题目大意:有一个鞋匠接了 n 双要修的鞋子, 修每双鞋需要 d 天,每推迟一天修将亏损 val 元,问按什么样的顺序修鞋可以保证损失最少,如果有多种情况输出字典序最小的
解题思路:因为鞋匠同时接到所有订单,起始罚款的日期相同,贪心fine/day ,就是鞋匠一天可以减少的最大损失,稳定排序即可
#include <cstdio>
#include <algorithm>
using namespace std;
struct Job {
int fine;
int time;
int order;
} job[1005];
bool cmp(Job x, Job y) {
return x.fine * y.time > y.fine * x.time;
}
int main() {
int t, n;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d%d", &job[i].time, &job[i].fine);
job[i].order = i + 1;
}
sort(job, job + n, cmp);
for (int i = 0; i < n; i++)
printf("%d%s", job[i].order, i == n - 1 ? t ? "\n\n" : "\n" : " ");
}
return 0;
}
解题思路:因为鞋匠同时接到所有订单,起始罚款的日期相同,贪心fine/day ,就是鞋匠一天可以减少的最大损失,稳定排序即可
#include <cstdio>
#include <algorithm>
using namespace std;
struct Job {
int fine;
int time;
int order;
} job[1005];
bool cmp(Job x, Job y) {
return x.fine * y.time > y.fine * x.time;
}
int main() {
int t, n;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d%d", &job[i].time, &job[i].fine);
job[i].order = i + 1;
}
sort(job, job + n, cmp);
for (int i = 0; i < n; i++)
printf("%d%s", job[i].order, i == n - 1 ? t ? "\n\n" : "\n" : " ");
}
return 0;
}
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