【贪心算法】Cross the river

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#Cross #rivier #贪心算法

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本文介绍了一种算法，用于解决一群人过河的问题，通过优化船只的载重限制和人数，计算出最少需要多少船只才能让所有人过河。输入包括人数、船只载重限制和每个人的重量，输出是最少船只数。

Problem description:

Some people want to cross a river by boat. Each person has a weight, and each boat can carry a maximum weight of limit. Each boat carries at most 2 people at the same time, provided the sum of the weight of those people is at most limit. Return the minimum number of boats to carry every given person. Note that it is guaranteed each person can be carried by a boat.

Input:

The input will be 3 lines. The first line is a single integer n (1 <= n <= 50000), which means the number of people. The second line is a single integer l (1 <= l <= 30000), represents the weight limit. And the last line contains n integers p_1, p_2, …, p_n (1 <= p_i <= l), separated by spaces, that represent people’s weights.

Output:

The output will consist of one line containing one integer, which represents the minimum number of boats.

Sample Input:

4
3
3 2 2 1

Sample Output:

原题地址：

UOJ#31

#include <iostream>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
class Solution {
public:
    int minimumBoats(int n, int limit, int persons[]) {
        sort(persons, persons + n);  // 从小到大排序
        // i从左到右,j从右到左
        for(int i = 0,j = n - 1;i <= j;j--){
            sum = persons[i] + persons[j];
            if(sum <= limit){
                numBoat++;
                i++;
            }else
                numBoat++;
        }
        return numBoat;
    }
private:
    int sum, numBoat = 0;  // 船的个数
};
int main(){
    int n, limit;
    cin >> n >> limit;
    int persons[n];
    for(int i = 0;i < n;i++)
        scanf("%d", &persons[i]);
    Solution sol;
    cout<<sol.minimumBoats(n, limit, persons)<<endl;
    return 0;
}