Codeforces 570 A. Elections

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原文链接：https://yq.aliyun.com/articles/15429

本文介绍了一种特殊的选举算法，该算法通过两轮投票决定胜者。首先，在每个城市中选取得票最高的候选人，若票数相同则编号较小者胜出。其次，从各城市的胜者中选出获胜次数最多的人为最终赢家，若次数相同则编号较小者获胜。

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                                           ***A. Elections***

The country of Byalechinsk is running elections involving n candidates. The country consists of m cities. We know how many people in each city voted for each candidate.

The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index.

At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index.

Determine who will win the elections.

Input
The first line of the input contains two integers n, m (1 ≤ n, m ≤ 100) — the number of candidates and of cities, respectively.

Each of the next m lines contains n non-negative integers, the j-th number in the i-th line aij (1 ≤ j ≤ n, 1 ≤ i ≤ m, 0 ≤ aij ≤ 109) denotes the number of votes for candidate j in city i.

It is guaranteed that the total number of people in all the cities does not exceed 109.

Output
Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one.

题意：
给出n行m列，代表第i个城市对第j的选手的票数data[i][j]，选举分为两次，第一次，每个城市选出这个城市得票最高的选手成为胜利者，如果票数一样，选编号最小的，然后第二轮，选择这些人里面在各个城市被选的胜利次数最多的选手作为最终胜利者，如果次数一样，选编号最小的。

直接上代码:

/*
Date : 2015-8-26 晚上
Author : ITAK

Motto :

今日的我要超越昨日的我，明日的我要胜过今日的我；
以创作出更好的代码为目标，不断地超越自己。
*/
#include <iostream>
#include <cstring>
using namespace std;
int data[105][105];
int ans[105];
int main()
{
    int m, n, j, i, ind;
    while(cin>>m>>n)
    {
        memset(ans, 0, sizeof(ans));
        for(i=0; i<n; i++)
        {
            int Max = -999;
            for(j=0; j<m; j++)
            {
                cin>>data[i][j];
                if(data[i][j] > Max)
                {
                     Max = data[i][j];
                     ind = j;
                }
            }
            ans[ind]++;
        }
        int Max = -9999;
        for(i=0; i<m; i++)
            if(ans[i] > Max)
            {
                Max = ans[i];
                ind = i;
            }
        cout<<ind+1<<endl;
    }
    return 0;
}
/**
input
3 3
1 2 3
2 3 1
1 2 1

output
2

input
3 4
10 10 3
5 1 6
2 2 2
1 5 7

output
1
**/