意甲冠軍 ?參加大ACM競(jìng)爭(zhēng)是非常回落喬布斯 ?每一個(gè)工作都有截止日期 ? 未完成必要的期限結(jié)束的期限內(nèi)扣除相應(yīng)的積分 ? 求點(diǎn)扣除的最低數(shù)量

把全部作業(yè)按扣分大小從大到小排序 ?然后就貪阿 ?能完畢前面的就完畢前面的 ?實(shí)在不能的就扣分吧~

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 1005; int dli[N], red[N], k[N], cas, ans, n; bool vis[N];bool cmp (int i, int j) {return red[i] > red[j]; }int main() {scanf ("%d", &cas);while (cas--){ans = 0;memset (vis, 0, sizeof (vis));scanf ("%d", &n);for (int i = 1; i <= n; ++i)scanf ("%d", &dli[i]), k[i] = i;for (int j = 1; j <= n; ++j)scanf ("%d", &red[j]);sort (k + 1, k + n + 1, cmp);for (int i = 1, j; i <= n; ++i){for (j = dli[k[i]]; j >= 1; --j)if (!vis[j]){vis[j] = 1;break;}if (j == 0) ans += red[k[i]];}printf ("%d\n", ans);}return 0; }<span style="font-family:Comic Sans MS;"> </span>

Doing Homework again

Problem Description Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score. ?
Input The input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=1000) which indicate the number of homework.. Then 2 lines follow. The first line contains N integers that indicate the deadlines of the subjects, and the next line contains N integers that indicate the reduced scores.
?
Output For each test case, you should output the smallest total reduced score, one line per test case.
?
Sample Input 3 3 3 3 3 10 5 1 3 1 3 1 6 2 3 7 1 4 6 4 2 4 3 3 2 1 7 6 5 4 ?
Sample Output 0 3 5 ?

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