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hihocoder-Week175-Robots Crossing River

时间:2017-11-06 20:04:30      阅读:201      评论:0      收藏:0      [点我收藏+]

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hihocoder-Week175-Robots Crossing River

Robots Crossing River

时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述

Three kinds of robots want to move from Location A to Location B and then from Location B to Location C by boat.

The only one boat between A and B and only one between B and C. Moving from A to B (and vise versa) takes 2 hours with robots on the boat. Moving from B to C (and vice versa) takes 4 hours. Without robots on the boat the time can be reduced by half. The boat between A and B starts at time 0 moving from A to B. And the other boat starts 2 hours later moving from B to C.

You may assume that embarking and disembarking takes no time for robots.

There are some limits:

1. Each boat can take 20 robots at most.

2. On each boat if there are more than 15 robots, no single kind of robots can exceed 50% of the total amount of robots on that boat.

3. At most 35 robots are allowed to be stranded at B. If a robot goes on his journey to C as soon as he arrives at B he is not considered stranded at B.

Given the number of three kinds robots what is the minimum hours to take them from A to C? 

输入

Three integers XY and Z denoting the number of the three kinds of robots. (0 ≤ XY and Z ≤ 1000)

输出

The minimum hours.

样例输入
40 4 4 
样例输出
24

 

 

使用贪心。

首先这是一道非常考验解决问题能力的题目。

(参考hihocoder的题目分析才做出来的, from: http://hihocoder.com/discuss/question/5025 )

首先需要看出整个流程的瓶颈完全在B-C这一段,换句话说我们只需求出所有机器人从B到C的最少时间,再加上2小时就是答案。事实上这个时间恰好等于把所有机器人直接从B运到C最少需要的船次x6。

如果没有“一船超过15个机器人则每种机器人不能超过半数”的限制,我们只需要20/船运走即可,最少船次是ceil((X+Y+Z)/20)。

由于有上面的限制,我们需要仔细讨论一下XYZ的相对大小。不妨设X >= Y >= Z,同时我们称三种机器人也为X、Y、Z类。

1、如果X <= Y + Z,那么我们仍然可以20/船运走,同时所有船都没有机器人超过半数。

这种情况最少船次仍然是是ceil((X+Y+Z)/20)。

2、 如果X > Y + Z,这时我们没办法使所有船都载20机器人。但是我们当然希望能尽量派出载20机器人的船。于是有如下贪心策略:

1) 首先尽量10个X类和10个非X类组成一船,派出若干船直到非X类不足10个机器人。 
2) 余下若干(不足10个)非X类机器人,配合尽可能多X类机器人组成一船。这里需要讨论余下的非X类机器人有多少个,不妨设为K。如果K不足8个,那么最多配合15-K个X类机器人,组成一船15个机器人;否则可以配合K个X类机器人,组成一船2K个机器人。 
3) 最后只剩下若干X类机器人,这些机器人只能15/船派出

 

 

 

#include <cstdio> 
#include <cstdlib> 

int cmp(const void *a, const void *b){
	return (*(int *)a - *(int *)b); 
}

int main(){

	int ans, num[3]; 

	while(scanf("%d %d %d", &num[0], &num[1], &num[2]) != EOF){
		qsort(num, 3, sizeof(num[0]), cmp); 

		ans = 0;  
		while(num[0] + num[1] + num[2] > 0){ 
			if( (num[0] + num[1]) >= num[2] ){
				ans += ( num[0] + num[1] + num[2] ) / 20; 
				if( (num[0] + num[1] + num[2]) % 20 != 0 ){
					++ans; 
				} 
				num[0] = num[1] = num[2] = 0; 
			}else{
				while(num[2] >= 10 && num[1] >= 10){
					++ans; 
					num[2] -= 10; 
					num[1] -= 10; 
					qsort(num, 3, sizeof(num[0]), cmp); 
				}
				if(num[2] >= 10){ 
					if( num[1] + num[0] >= 10 ){
						num[2] -= 10; 
						num[1] -= (10 - num[0]); 
						num[0] = 0; 
						++ans; 
					}else if(num[1] + num[0] >= 8){
						num[2] -= ( num[1] + num[0] ); 
						num[1] = num[0] = 0; 
						++ans;  
					}
				} 
				ans += ( num[2] + num[1] + num[0] ) / 15; 
				if( ( num[2] + num[1] + num[0] )%15 != 0 ){
					++ans; 
				}
				break; 
			} 
		}
		ans = 4 * ans + 2*(ans - 1) + 2; 
		printf("%d\n", ans );
	} 
	return 0; 
}

  

 

hihocoder-Week175-Robots Crossing River

标签:参考   输出   解决   hat   var   rri   limits   大小   rsa   

原文地址:http://www.cnblogs.com/zhang-yd/p/7794671.html

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