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?
输入
Three integers X, Y and Z denoting the number of the three kinds of robots. (0 ≤ X, Y 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
", ans );
}
return 0;
}