【贪心算法】POJ-3262

最新推荐文章于 2020-04-17 14:40:12 发布

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最新推荐文章于 2020-04-17 14:40:12 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/CSLaker/p/7341263.html

本文介绍了一道关于贪心算法的经典问题：如何通过合理安排顺序来最小化被破坏花朵的数量。文章提供了完整的解析思路及C++实现代码，通过计算每头牛破坏单位时间内花朵数量的比例来进行排序，从而得出最优运输顺序。

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一、题目

Description

Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When he returned, he found to his horror that the cluster of cows was in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport each cow back to its own barn.

Each cow i is at a location that is Ti minutes (1 ≤ Ti ≤ 2,000,000) away from its own barn. Furthermore, while waiting for transport, she destroys Di (1 ≤ Di ≤ 100) flowers per minute. No matter how hard he tries, FJ can only transport one cow at a time back to her barn. Moving cow i to its barn requires 2 × Ti minutes (Ti to get there and Ti to return). FJ starts at the flower patch, transports the cow to its barn, and then walks back to the flowers, taking no extra time to get to the next cow that needs transport.

Write a program to determine the order in which FJ should pick up the cows so that the total number of flowers destroyed is minimized.

Input

Line 1: A single integer N
Lines 2..N+1: Each line contains two space-separated integers, Ti and Di, that describe a single cow's characteristics

Output

Line 1: A single integer that is the minimum number of destroyed flowers

Sample Input

6
3 1
2 5
2 3
3 2
4 1
1 6

Sample Output

二、思路&心得

贪心策略：用d/t的值进行选择，值大的数据先处理，以达到花费代价最小。
可以使用排序或则优先队列做，个人觉得排序比较简单直接。至于每一次数据处理时的代价计算，利用总代价反复减去对应数据的代价即可。
题目需注意是数据量较大，所有结果应采用long long或则__int64类型进行定义。

三、代码

#include<cstdio>
#include<algorithm>
using namespace std;
const int MAX_N = 100005; 

typedef pair<float, float> P;

int N, sum;

long long cost;

P a[MAX_N];

bool cmp(P a, P b) {
    return a.second / a.first > b.second / b.first;
}

void solve() {
    sum = 0, cost = 0;
    for (int i = 0; i < N; i ++) {
        scanf("%f %f", &a[i].first, &a[i].second);
        sum += a[i].second;
    }
    sort(a, a + N, cmp);
    for (int i = 0; i < N; i ++) {
        sum -= a[i].second;
        cost += 2 * a[i].first * sum;
    }
    printf("%lld\n", cost);
}

int main() {
    while (~scanf("%d", &N)) {
        solve();
    }
    return 0;
}

转载于:https://www.cnblogs.com/CSLaker/p/7341263.html