ACM学习历程——POJ 1700 Crossing River（贪心）

最新推荐文章于 2020-03-20 20:22:16 发布

weixin_30414305

最新推荐文章于 2020-03-20 20:22:16 发布

阅读量92

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/andyqsmart/p/4116342.html

本文探讨了如何通过策略优化，使一组人在限定时间内完成过河任务，关注于最小化总过河时间。

Description

A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must be arranged in order to row the boat back and forth so that all people may cross. Each person has a different rowing speed; the speed of a couple is determined by the speed of the slower one. Your job is to determine a strategy that minimizes the time for these people to get across.

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. The first line of each case contains N, and the second line contains N integers giving the time for each people to cross the river. Each case is preceded by a blank line. There won't be more than 1000 people and nobody takes more than 100 seconds to cross.

Output

For each test case, print a line containing the total number of seconds required for all the N people to cross the river.

Sample Input

1
4
1 2 5 10

Sample Output

17


这个题目第一反应是船回的时候，肯定让当前最快的人划回来，然后还考虑到，去的时候是按照最小速度的人，故让最慢的两个人过去，可能会减少时间。
于是，考虑两种贪心策略：
1、让最快的人来回载人过去。
2、先让最快的人和次快的人过去，然后让最快的人回来，让最慢和次慢的人过去，让对面最快的人回来。整体效果是最快和次快的人让最慢和次慢的人过去。
于是两者统一，就是让最慢和次慢的人每次渡河。比较两种策略的时间。
直到剩余两个或者三个人，就稍微考虑一下即可。另外不要忘了考虑初始只有一个人的情况。


代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <string>
#define inf 0x3fffffff
#define eps 1e-10

using namespace std;

int n, a[1005];

bool cmp(int x, int y)
{
    return x > y;
}

void Input()
{
    scanf("%d", &n);
    for (int i = 0; i < n; ++i)
        scanf("%d", &a[i]);
    sort(a, a+n, cmp);
}

int qt()
{
    if (n == 1 || n == 2)
        return a[0];
    int ans = 0;
    int top = 0, rear = n;
    while (rear - top > 3)
    {
        ans += min(a[rear-2]*2+a[rear-1]+a[top], a[top]+a[top+1]+2*a[rear-1]);
        top += 2;
    }
    if (rear - top == 2)
        ans += a[rear-2];
    else
        ans += a[top] + a[rear-1] + a[rear-2];
    return ans;
}

int main()
{
    //freopen("test.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = 0; times < T; ++times)
    {
        Input();
        printf("%d\n", qt());
    }
    return 0;
}

转载于:https://www.cnblogs.com/andyqsmart/p/4116342.html