Design T-Shirt

Design T-Shirt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4273    Accepted Submission(s): 2076

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

 

水题，结构题二级排序，在对编号进行逆序排序

 

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct {
    float sa;
    int nu;
}Node;
Node logo[];
int temp[];

int comp1(const void *a,const void *b)
{
    Node *p1,*p2;
    p1 = (Node *)a;
    p2 = (Node *)b;
    if(p1->sa!=p2->sa)
        return p2->sa-p1->sa;
    return p1->nu-p2->nu;
}

int comp2(const void *a,const void *b)
{
    return *(int*)b-*(int *)a;
}
void init(int n)
{
    int i;
    for(i =; i < n; i ++)
    {
        logo[i].sa = 0.;
        logo[i].nu = i+;
    }
}

int main()
{
    int i,j,n,m,k;
    float a;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        memset(temp,,sizeof(temp[]));
        init(m);
        for(i =;i < n; i ++)
        {
            for(j =; j < m; j ++)
            {
                scanf("%f",&a);
                logo[j].sa += a;
            }
        }
       qsort(logo,m,sizeof(logo[]),comp1);
       for(i =;i < k; i ++)
        temp[i] = logo[i].nu;
       qsort(temp,k,sizeof(temp[]),comp2);
       for(i =;i < k-; i ++)
        printf("%d ",temp[i]);
       printf("%d\n",temp[i]);
    }
    return;
}