Very Simple Problem

Very Simple Problem

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2535

Description

During a preparation of programming contest, its jury is usually faced with many difficult tasks. One of them is to select a problem simple enough to most, if not all, contestants to solve.

The difficulty here lies in diverse meanings of the term "simple" amongst the jury members. So, the jury uses the following procedure to reach a consensus: each member weights each proposed problem with a positive integer "complexity rating" (not necessarily different for different problems). The jury member calls "simplest" those problems that he gave the minimum complexity rating, and "hardest" those problems that he gave the maximum complexity rating.

The ratings received from all jury members are then compared, and a problem is declared as "very simple", if it was called as "simplest" by more than a half of the jury, and was called as "hardest" by nobody.

Input

The first line of input file contains integers N and P, the number of jury members and the number of problems. The following N lines contain P integers in range from 0 to 1000 each - the complexity ranks. 1 <= N, P <= 100

Output

Output file must contain an ordered list of problems called as "very simple", separated by spaces. If there are no such problems, output must contain a single integer 0 (zero).

Sample Input

4 4
1 1 1 2
5 900 21 40
10 10 9 10
3 4 3 5

Sample Output

3

/*
题意：有n位评委，给p个选手打分，让你找出very simple的选手，有这样一个标准就是给他打最低分的评委数量超过一半，并且没有评委给他打过最低分

初步思路：模拟
*/
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int n,p;
int a[][];
int cur[];//用来标记每个选手的最低分得票 cur[i]==-1表示这个选手的过最高分，那么他的积分就不做评价
int maxn,minn;
void init(){
    memset(cur,,sizeof cur);
}
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&p)!=EOF){
        init();
        for(int i=;i<n;i++){
            maxn=-;
            minn=;
            for(int j=;j<p;j++){
                scanf("%d",&a[i][j]);
                if(a[i][j]>maxn){
                    maxn=a[i][j];
                }
                if(a[i][j]<minn){
                    minn=a[i][j];
                }
            }
            for(int j=;j<p;j++){
                if(a[i][j]==maxn)
                    cur[j]=-;
                else if(a[i][j]==minn){
                    if(cur[j]==-)
                        continue;
                    else
                        cur[j]++;
                }

            }
        }
        // for(int i=0;i<n;i++){
        //     cout<<cur[i]<<" ";
        // }
        // cout<<endl;
        bool flag=false;
        for(int i=;i<n;i++){
            if(cur[i]>n/){
                if(flag){
                    printf(" %d",i+);
                }else{
                    printf("%d",i+);
                    flag=true;
                }
            }
        }
        if(flag==false)
            printf("");
        printf("\n");
    }
    return ;
}