Design T-Shirt
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7637 Accepted Submission(s):
3616
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
Author
CHEN, Yue
Source
Recommend
简单排序，注意题意，以及输入的测试数据有小数点，所以是double型。
题意：给出n个人对m件T恤的满意度，按编号降序输出这m件T恤中综合满意度前k大的。多组输入，每组用例第一行为三个整数n,m,k,之后一个n*m矩阵表示n个人对m件T恤的满意度，以文件尾结束输入。对于每组用例，输出综合满意度前k大的(按编号降序输出) 。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 struct node 7 { 8 int t; 9 double w; 10 } ss[10005]; 11 bool cmp1(node a,node b) 12 { 13 return a.w>b.w; 14 } 15 bool cmp2(node a,node b) 16 { 17 return a.t>b.t; 18 } 19 int main() 20 { 21 int n,m,i,j,k; 22 double s; 23 while(~scanf("%d%d%d",&n,&m,&k)) 24 { 25 memset(ss,0,sizeof(ss)); 26 for(j=1; j<=n; j++) 27 { 28 for(i=1; i<=m; i++) 29 { 30 ss[i].t=i; 31 scanf("%lf",&s); 32 ss[i].w+=s; //每件衣服每个人的满意度累加 33 } 34 } 35 36 sort(ss+1,ss+m+1,cmp1); //满意度从大到小排序 37 sort(ss+1,ss+k+1,cmp2); //输出序号从大到小排序 38 for(i=1; i<=k; i++) 39 { 40 if(i!=1) printf(" "); 41 printf("%d",ss[i].t); 42 } 43 printf(" "); 44 } 45 return 0; 46 }