hihocoder-Week175-Robots Crossing River

Robots Crossing River

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?

輸入

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

輸出

The minimum hours.

樣例輸入

40 4 4 

樣例輸出

24

使用貪心。

首先這是一道非常考驗解決問題能力的題目。

（參考hihocoder的題目分析才做出來的， from： http://hihocoder.com/discuss/question/5025 ）

首先需要看出整個流程的瓶頸完全在B-C這一段，換句話說我們只需求出所有機器人從B到C的最少時間，再加上2小時就是答案。事實上這個時間恰好等於把所有機器人直接從B運到C最少需要的船次x6。

如果沒有“一船超過15個機器人則每種機器人不能超過半數”的限制，我們只需要20/船運走即可，最少船次是ceil((X+Y+Z)/20)。

由於有上面的限制，我們需要仔細討論一下XYZ的相對大小。不妨設X >= Y >= Z，同時我們稱三種機器人也為X、Y、Z類。

1、如果X <= Y + Z，那麽我們仍然可以20/船運走，同時所有船都沒有機器人超過半數。

這種情況最少船次仍然是是ceil((X+Y+Z)/20)。

2、 如果X > Y + Z，這時我們沒辦法使所有船都載20機器人。但是我們當然希望能盡量派出載20機器人的船。於是有如下貪心策略：

1) 首先盡量10個X類和10個非X類組成一船，派出若幹船直到非X類不足10個機器人。
2) 余下若幹(不足10個)非X類機器人，配合盡可能多X類機器人組成一船。這裏需要討論余下的非X類機器人有多少個，不妨設為K。如果K不足8個，那麽最多配合15-K個X類機器人，組成一船15個機器人；否則可以配合K個X類機器人，組成一船2K個機器人。
3) 最後只剩下若幹X類機器人，這些機器人只能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