HDU1789 贪心
Doing Homework again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1221 Accepted Submission(s): 715
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
想了好久 , 原来是贪心 , 把所有作业按照截止日期从小到大排序 , 然后把所有截止日期也提取到数组 d[] 排序 , 然后从后到前进行贪心选择 . 找出所有截止日期大于或等于 d[i] 的作业 , 如果数量比 d[i]-d[i-1] 大 , 那么就把这些作业按照扣分从小到大排序 , 除去扣分最多的 (d[i]-d[i-1]) 个 , 剩下的放到后面重复进行贪心选择 .
#include<stdio.h> #include<string.h> #include<algorithm> using std::sort; #define N 1005 struct Homework { int d,s; bool operator < (const Homework a) const { if (d<a.d) return 1; return 0; } }h[N]; bool cmp(Homework a, Homework b) { if (a.s<b.s) return 1; return 0; } int main() { int n,i,j,total,T,k,sum,d[N]; scanf("%d",&T); while (T--) { scanf("%d",&n); for (i=1;i<=n;++i) scanf("%d",&h[i].d); for (i=1;i<=n;++i) scanf("%d",&h[i].s); sort(h+1,h+n+1); total=0; h[0].s=0; memset(d,0,sizeof(d)); for (i=1;i<=n;++i) { if (h[i].d!=d[total]) d[++total]=h[i].d; } k=total; i=j=n; while (k) { while (h[i].d==d[k]) i--; if (j-i<=d[k]-d[k-1]) { j=i; } else { sort(h+i+1,h+j+1,cmp); j-=(d[k]-d[k-1]); } k--; } sum=0; for (k=i;k<=j;++k) sum+=h[k].s; printf("%d/n",sum); } return 0; }