hdu 1789 - Doing Homework again
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题目:有很多科目的作业要做,每个作业有个上交的截止时间,每个作业不交会扣一定的分数,
每个作业需要有1天完成,求最小的扣分数。
分析:贪心。每次找到当前剩余的作业中分数最高的,找到小于它的截止日期的未被安排的最大时间,
对这个时间进行占用,寻找下一个作业即可。
(如果截止日期相同,一定安排分数大的;如果截止日期不同,出现重叠情况,也是先安排大的好)
(如果分数相同,先安排截止日期靠前的,它的选择空间小,这样可能得到更好的安排情况)
说明:UVa打不开,先刷hdu吧,╮(╯▽╰)╭。
#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;typedef struct work{int deadline,score;}work;work W[1001];bool day[1001];bool cmp(work a, work b){if (a.score != b.score)return a.score > b.score;return a.deadline < b.deadline;}int main(){int n,m;while (~scanf("%d",&n)) while (n --) {scanf("%d",&m);for (int i = 0 ; i < m ; ++ i)scanf("%d",&W[i].deadline);for (int i = 0 ; i < m ; ++ i)scanf("%d",&W[i].score);sort(W, W+m, cmp);memset(day, 0, sizeof(day));int sum = 0;for (int j,i = 0 ; i < m ; ++ i) {for (j = W[i].deadline ; j ; -- j)if (!day[j]) {day[j] = 1;break;}if (!j) sum += W[i].score;}printf("%d\n",sum);}return 0;}
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