zoukankan      html  css  js  c++  java
  • HDU 1031.Design T-Shirt【结构体二次排序】【8月21】

    Design T-Shirt

    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
    n个人,m件衣服,输出评价最高的前k件,输出时,评价同样的标号大的先输出。一定要二次排,由于可能所有的评价一样高,仅仅排一次输出就会有错误。

    代码例如以下:

    #include<cstdio>
    #include<algorithm>
    using namespace std;
    struct ss{
        double v;
        int x;
    };
    bool cmp1(ss x,ss y){
        if(x.v>y.v) return true;
        else return false;
    }
    bool cmp2(ss x,ss y){
        if(x.x>y.x) return true;
        else return false;
    }
    int main(){
        int n,m,k;
        double t;
        ss f[1010];
        while(scanf("%d%d%d",&n,&m,&k)==3){
            for(int i=0;i<m;i++)
                f[i].v=0;
            while(n--)
            for(int i=0;i<m;i++){
                scanf("%lf",&t);
                f[i].v+=t;
                f[i].x=i+1;
            }
            sort(f,f+m,cmp1);
            sort(f,f+k,cmp2);//二次排。仅仅排前k项
            for(int i=0;i<k;i++){
                if(i!=0) printf(" ");
                printf("%d",f[i].x);
            }
            printf("
    ");
        }
        return 0;
    }
    

  • 相关阅读:
    我的作业
    代码练习
    prority_queue 的用法 实例
    最短路问题专题
    键值对 Intent
    P103 任意两点之间的最短路问题 Floyd_warshall算法
    第6届山东省ACM省赛总结
    HDU 3247 Resource Archiver[AC自动机+最短路+dp]
    POJ 2778 DNA Sequence [AC自动机 + 矩阵快速幂]
    HDU 2896 病毒侵袭 [AC自动机]
  • 原文地址:https://www.cnblogs.com/yangykaifa/p/7182927.html
Copyright © 2011-2022 走看看