C++ 投票选举问题： 最少劝退候选人数量

首先，我们需要统计每个候选人的得票数。遍历每个县的投票顺序，将每个候选人的得票数加一，直到出现前一个候选人退出选举的情况。

然后，我们可以计算出当前最高得票数的候选人编号。如果最高得票数的候选人编号等于k，那么他就是获胜者，不需要任何候选人退出。否则，我们需要计算至少让多少候选人退出，才能保证候选人k获胜。我们只需要将其他候选人的得票数归零，然后重新统计最高得票数的候选人编号，直到最高得票数的候选人编号等于k。

下面是一个可能的实现：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n, m, k;
    cin >> n >> m >> k;

    vector<vector<int>> votes(n, vector<int>(m));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> votes[i][j];
        }
    }

    vector<int> counts(m, 0);
    vector<bool> eliminated(m, false);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            int candidate = votes[i][j] - 1;
            if (!eliminated[candidate]) {
                counts[candidate]++;
                break;
            }
        }
    }

    int maxCount = *max_element(counts.begin(), counts.end());
    int winner = max_element(counts.begin(), counts.end()) - counts.begin() + 1;
    cout << winner << endl;

    int minEliminations = 0;
    while (winner != k) {
        int minCount = maxCount;
        for (int i = 0; i < m; i++) {
            if (!eliminated[i] && counts[i] < minCount) {
                minCount = counts[i];
            }
        }
        for (int i = 0; i < m; i++) {
            if (counts[i] == minCount) {
                eliminated[i] = true;
                minEliminations++;
            }
        }
        counts.assign(m, 0);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int candidate = votes[i][j] - 1;
                if (!eliminated[candidate]) {
                    counts[candidate]++;
                    break;
                }
            }
        }
        maxCount = *max_element(counts.begin(), counts.end());
        winner = max_element(counts.begin(), counts.end()) - counts.begin() + 1;
    }

    cout << minEliminations << endl;

    return 0;
}