Design T-Shirt 分类: HDU 2015-06-26 11:58 7人阅读 评论(0) 收藏
2015-06-26 11:58
399 查看
Design T-Shirt
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6744 Accepted Submission(s): 3167
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
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
[align=left]Sample Output[/align]
6 5 3 1 2 1#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <string> #include <stack> #include <algorithm> using namespace std; const int Max=1100000; struct ELE { int num; double sum; } E[1100]; int a[1100]; bool cmp1(ELE a,ELE b) { if(a.sum>b.sum||(a.sum==b.sum&&a.num<b.num)) return true; return false; } bool cmp2(int a,int b) { return a>b; } int main() { int n,m,k; double data; while(~scanf("%d %d %d",&n,&m,&k)) { for(int i=0; i<n; i++) { for(int j=0; j<m; j++) { scanf("%lf",&data); if(i) { E[j].sum+=data; } else { E[j].num=j+1; E[j].sum=data; } } } sort(E,E+m,cmp1); for(int i=0;i<k;i++) { a[i]=E[i].num; } sort(a,a+k,cmp2); for(int i=0;i<k;i++) { if(i) cout<<" "; cout<<a[i]; } cout<<endl; } return 0; }
相关文章推荐
- Win8.1无法关机问题解决方法
- 图解MBR分区无损转换GPT分区+UEFI引导安装WIN8.1
- 图解MBR分区无损转换GPT分区+UEFI引导安装WIN8.1
- 用itext生成pdf报表上篇
- 剑指offer 例题
- 《开源框架那点事儿19》:特斯拉建“桩”与开源的生命力
- D题 Codeforces 325B Stadium And Games
- 图解MBR分区无损转换GPT分区+UEFI引导安装WIN8.1
- 网络编程API-下 (I/O复用函数)
- 图解MBR分区无损转换GPT分区+UEFI引导安装WIN8.1 分类: Windows-嵌入式 生活百科 2015-06-26 11:57 475人阅读 评论(2) 收藏
- Redis有序集内部实现原理分析(二)
- HTML 标签 <body>
- Maven详解之仓库------本地仓库、远程仓库
- 贝塞尔曲线 (Bézier curve) 理论及绘制方法
- matlab纵波动画
- Android之activity总结
- cocos2d-x 3.0rc1 创建project
- POJ 1741 Tree (树上点分治)(楼教主男人八题之一)
- 事件委托
- 更改Android应用程序的图标