hdoj 1031 Design T-Shirt



http://acm.hdu.edu.cn/showproblem.php?pid=1031

Design T-Shirt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5852    Accepted Submission(s): 2730

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 4
2 2.5 5 1 3 4
5 1 3.5 2 2 2
1 1 1 1 1 10
3 3 2
1 2 3
2 3 1
3 1 2
  
  

 

Sample Output

  
  

   
   6 5 3 1
2 1
  
  
  
  


  
  
  
  

   
   题目本身不难，是结构体排序，只是题意有点纠结，让N个人(编号从1~N)对M件衣服打分，输出前K件打分高的衣服序号，而且前K件也要按序号从大到小输出。
  
  
  
  

   
   先将每一联的分数加起来，再排序，然后再对前K个分数按衣服序号从大到小排序。
  
  
  
  

   
   
   
   
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAX 1000
struct Sth
{
	int N;
	double sum;
}num[MAX];
double a[MAX][MAX];
int cmp_M(const void *a,const void *b)//对num[].sum从大到小排序 
{
	return ((Sth *)b)->sum - ((Sth *)a)->sum;
}
int cmp_K(const void *a,const void *b)//对前K个最大的按num[].N的大小从大到小排序 
{
	return ((Sth *)b)->N - ((Sth *)a)->N;
}
int main()
{
	int N,M,K,i,j;
	while(scanf("%d %d %d",&N,&M,&K)!=EOF)
	{
		for(i=0;i<MAX;i++)
		memset(a[i],0,sizeof(a[i]));
		for(i=0;i<N;i++)
		{
			for(j=0;j<M;j++)
			scanf("%lf",&a[i][j]);
		}
		for(i=0;i<M;i++)
		{
			num[i].N=i+1;
			for(j=0,num[i].sum=0;j<N;j++)
			num[i].sum+=a[j][i];
		}
		qsort(num,M,sizeof(num[0]),cmp_M);
		
		
		qsort(num,K,sizeof(num[0]),cmp_K);
		for(i=0;i<K;i++)
		{
			printf("%d",num[i].N);
			if(i!=K-1)
			printf(" ");
		}
		printf("\n");
	}
	return 0;
}