POJ 1700 - Crossing River

最新推荐文章于 2021-02-08 10:56:52 发布

Freenm

最新推荐文章于 2021-02-08 10:56:52 发布

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分类专栏： ACM - 贪心文章标签： poj 贪心

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本文介绍了一种关于多人利用一艘船过河的最优策略问题，通过贪心算法确定最少时间安排，涉及人员排序及不同组合方式的比较。

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13982		Accepted: 5349

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

引用来自http://www.w2bc.com/Article/18634的思路，按照自己的习惯做了些修改：

贪心思想（一般都是先排序）
每次从此岸到对岸移动的两个人要么这两个人中有一个是时间最快的那个人，要么这两个人到达对岸后再也不回来。即：要么最快+最慢（最快回来换人），要么最慢+次慢（不回来）。
1.对N个人过河时间从小到大排序。p[i];
2.分情况讨论：
⑴当n = 1，直接过河。sum = p[1]
(2)当n = 2，直接过河。 sum = p[2]
(3)当n = 3，无论怎么过河， sum = p[1] + p[2] + p[3]
(4)当n >= 4，设从小到大排序后位a,b,……,c,d，大于4个人，a,b是最小的两个人，c,d是最大的两个人，目标就是把最大的两个人送过去。就要牺牲最小的。
用最小的来送：A=d + a + c + a = 2a + c + d = 2*p[1] + p[end-1] + p[end];(a,d过去，a回来，a,c过去，a回来）
两小送两大：B= b + b + d + a = a + 2b + d = 2*p[1] + p[end-1] + p[end];(a,b过去，b回来，c,d过去，a回来）
循环：sum = min(A,B)，直到n <= 3 时候结束。

刚开始蠢，用函数递归调用，果断超时……

#include<cstdio>
#include<algorithm>
using namespace std;
int p[1003];
int min(int a,int b) {return a<b?a:b;}
int time_sum(const int p[],int end)
{
	if(end == 3) return p[1]+p[2]+p[3];
	if(end == 2) return p[2];
	if(end == 1) return p[1];
	int time_1= 2*p[1] + p[end-1] + p[end] + time_sum(p,end-2);
	int time_2= 2*p[2] + p[end] + p[1] + time_sum(p,end-2);
	return min(time_1,time_2);
}
int main()
{
	int t,n,time;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			scanf("%d",&p[i]);
		}
		sort(p+1,p+n+1);
		if(n == 1){
			time=p[1];
		}
		else if(n == 2){
			time=p[2];
		}
		else if(n == 3){
			time=p[1]+p[2]+p[3];
		}
		else if(n >= 4){
			time=time_sum(p,n);
		}
		printf("%d\n",time);
	}
}

后来改了下就好了：

#include<cstdio>
#include<algorithm>
using namespace std;
int p[1003];
int min(int a,int b) {return a<b?a:b;}
int main()
{
	int t,n,time;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			scanf("%d",&p[i]);
		}
		sort(p+1,p+n+1);
		if(n == 1){
			time=p[1];
		}
		else if(n == 2){
			time=p[2];
		}
		else if(n == 3){
			time=p[1]+p[2]+p[3];
		}
		else if(n >= 4){
			time=0;
			int end=n;
			while(end >= 4){
				time+=min( 2*p[1] + p[end-1] + p[end] , 2*p[2] + p[end] + p[1] );
				end-=2;
			}
			if(end == 3) time+=p[1]+p[2]+p[3];
			else if(end == 2) time+=p[2];
			else if(end == 3) time+=p[1];
		}
		printf("%d\n",time);
	}
}