1056. Mice and Rice (25)

时间限制

30 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

========

Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse.

First the playing order is randomly decided for N_P programmers. Then every N_G programmers are grouped in a match. The fattest mouse in a group wins and enters the next turn. All the losers in this turn are ranked the same. Every N_Gwinners are then grouped in the next match until a final winner is determined.
For the sake of simplicity, assume that the weight of each mouse is fixed once the programmer submits his/her code. Given the weights of all the mice and the initial playing order, you are supposed to output the ranks for the programmers.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers: N_P and N_G (<= 1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than N_G mice at the end of the player’s list, then all the mice left will be put into the last group. The second line contains N_P distinct non-negative numbers W_i (i=0,…N_P-1) where each W_iis the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,…N_P-1 (assume that the programmers are numbered from 0 to N_P-1). All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the final ranks in a line. The i-th number is the rank of the i-th programmer, and all the numbers must be separated by a space, with no extra space at the end of the line.
Sample Input:

11 3
25 18 0 46 37 3 19 22 57 56 10
6 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

===================================
这个题目主要的难度在时间限制，30ms看起来很吓人，但是其实问题的规模不算大…最多才1000只老鼠嘛！
主要思路就是矩阵数组能搞多大搞多大，能用静态用静态，因为内存范围太宽…
为了加快程序速度，在对组内的老鼠排序的时候，显然不能使用标准库里面的算法，就算快速排序也有O(nlog2n)的时间复杂度，这边只要选一只最肥的就行，用最笨的冒泡或者插入排序做一次就行了，O(n)的时间复杂度。
另外第二个难度可能在比赛结果的存储上，由于时间限制紧，我就直接弄了个1000*1000的矩阵存结果，第一行存的是第一轮比赛的结果，把没有晋级的老鼠们的序号存在第一行的各个列中，用-1作为本列结束标志…这样以此类推，当某行第一个元素就是-1的时候，所有肥老鼠就比完赛了（当然这个首列-1的设置在后面的排名中没起什么结果，因为有个计数器已经记下一共比了多少轮了）
另外，记得cin,cout用scanf,printf替换，不然读个数列写个结果估计就能让程序超时了。这个可能是PAT设计的不合理的地方，应该分程序语言来限定时间，30ms这么搞让写Java的情何以堪？？…
别的没啥了，自己的代码5个测试点都是0ms..内存消耗或许大了点，哈哈
====================================

#include 
using namespace std;
const int MAX_ELEM_SIZE = 1000;
int Player_values[MAX_ELEM_SIZE];
int Player_result[MAX_ELEM_SIZE][MAX_ELEM_SIZE] = {0}; //比赛结果,0表示还没有出最后成绩
int contest_size = 0; //每场比赛的总人数
int stadium[MAX_ELEM_SIZE]; //赛场,存储序号
int sequence[MAX_ELEM_SIZE];//比赛顺序，没有资格参赛的设置为-1
int ranks[MAX_ELEM_SIZE];
int N_P, N_G;
int main()
{
	scanf("%d%d", &N_P, &N_G);
	for(int i=0; i 0)
	{
		//一轮比赛一个循环
		int max_seq=-1;
		int max_val=-1;
		int true_count = 0;
		int loser_count = 0;
		for(int i=0; i max_val)
			{
				if(contest_size == 1)
				{
					//he is the winner!
					Player_result[turn][loser_count++] = sequence[i];
					contest_size = 0;
					continue;
				}
				max_val = Player_values[ sequence[i] ];
				if( max_seq != -1)
				{
					//把原来的人标记为Loser
					Player_result[turn][loser_count++] = sequence[max_seq];
					contest_size --;
					sequence[max_seq] = -1;
				}
				max_seq = i;
			}
			else
			{
				//loser
				Player_result[turn][loser_count++] = sequence[i];
				contest_size -- ;
				sequence[i] = -1;
			}
			true_count ++;
		}
		Player_result[turn][loser_count] = -1; //终止这一行！
		turn++;
	}
	Player_result[turn][0] = -1;
	//开始分配名次
	int rank = 1;
	int r_count = 1;
	while(turn --)
	{
		int col = 0;
		rank = r_count;
		while(true)
		{
			if( Player_result[turn][col] == -1)
				break;
			ranks[ Player_result[turn][col] ] = rank;
			r_count++;
			col ++;
		}
	}
	for(int i=0; i

Leave a Reply Cancel reply