Design T-Shirt（sort）

Design T-Shirt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10083    Accepted Submission(s): 4807

Problem Description

Soon after he decided to design a T-shirt for our Algorithm Board on Free-City BBS, XKA found that he was trapped by all kinds of suggestions from everyone on the board. It is indeed a mission-impossible to have everybody perfectly satisfied. So he took a poll to collect people's opinions. Here are what he obtained: N people voted for M design elements (such as the ACM-ICPC logo, big names in computer science, well-known graphs, etc.). Everyone assigned each element a number of satisfaction. However, XKA can only put K (<=M) elements into his design. He needs you to pick for him the K elements such that the total number of satisfaction is maximized.

 

Input

The input consists of multiple test cases. For each case, the first line contains three positive integers N, M and K where N is the number of people, M is the number of design elements, and K is the number of elements XKA will put into his design. Then N lines follow, each contains M numbers. The j-th number in the i-th line represents the i-th person's satisfaction on the j-th element.

 

Output

For each test case, print in one line the indices of the K elements you would suggest XKA to take into consideration so that the total number of satisfaction is maximized. If there are more than one solutions, you must output the one with minimal indices. The indices start from 1 and must be printed in non-increasing order. There must be exactly one space between two adjacent indices, and no extra space at the end of the line.

 

Sample Input

3 6 42 2.5 5 1 3 45 1 3.5 2 2 21 1 1 1 1 103 3 21 2 32 3 13 1 2

 

Sample Output

6 5 3 12 1

 

Author

/*求出前k个最大的满意度，按照降序排列，如果满意度相同则选择下标小的输出。用结构体存储id，和满意度。用sort排序按总满意度（若相同时，按id升序排序）降序排序一次。然后将前k个id存到整形数组，再用sort降序排序。 */#include<stdio.h>#include<algorithm>using namespace std; struct Node{int id;double  toal;};int cmp2(int a,int b){return a>b;}int cmp(Node a,Node b){if(a.toal==b.toal)return a.id<b.id;else return a.toal>b.toal;}int main(){int n,k;int m;while(scanf("%d%d%d",&n,&m,&k)!=EOF){Node st[1000];while(n--){double tt;for(int i=1;i<=m;i++){ st[i].id=i; scanf("%lf",&tt); st[i].toal+=tt;}}sort(st+1,st+m+1,cmp);/*for(int i=1;i<=m;i++){printf("%d %lf\n",st[i].id,st[i].toal);}*/int str[k]={0};int j=0;for(int i=1;i<=k;i++){str[j++]=st[i].id;} sort(str,str+k,greater<int>());for(int i=0;i<k-1;i++){printf("%d ",str[i]);}printf("%d\n",str[k-1]);//清空 for(int i=1;i<=m;i++){st[i].toal=0;st[i].id=0;}}}