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. InputThe 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. OutputFor 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 題意：ACM大佬比賽回來但無奈作業(yè)都沒寫完，現(xiàn)在給你他的每門課程的deadline和如果不寫完所扣的分?jǐn)?shù)。現(xiàn)在要求你這樣安排才能使扣分最少。輸出最少的分?jǐn)?shù)。 思路：貪心算法。先把分?jǐn)?shù)按從大到小排序，然后將最大的放在它的最后的期限做。依次來如果deadline有作業(yè)可做，就往前進(jìn)一個直到第一天為止。 AC代碼：
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct homework{ int dl; int score; bool flge;} hw[1005];bool vis[1005];bool cmp(homework a,homework b){ return a.score>b.score;}int main(){ int T; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&hw[i].dl); } for(int i=0;i<n;i++) { scanf("%d",&hw[i].score); } memset(vis,0,sizeof(vis)); for(int i=0;i<n;i++) { hw[i].flge=0; } sort(hw,hw+n,cmp); for(int i=0;i<n;i++) { for(int j=hw[i].dl;j>0;j--) { if(vis[j]==0) { vis[j]=1; hw[i].flge=1; break; } } } int sum=0; for(int i=0;i<n;i++) { if(hw[i].flge==0) { sum+=hw[i].score; } }新聞熱點(diǎn)
疑難解答
