1056. Mice and Rice (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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_G winners 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_i is 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:

11 325 18 0 46 37 3 19 22 57 56 106 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

解析:这道题对于竞技规则不了解的我真的好费劲, 查阅了网友的解释,题意总结如下:

有n个老鼠，第一行给出n个老鼠的重量，第二行给出他们的顺序。
1.每一轮分成若干组，每组m个老鼠，不能整除的多余的作为最后一组。
2.每组重量最大的进入下一轮。
让你给出每只老鼠最后的排名。
很简单，用两个数组模拟一下即可
order1存储进入当前一轮老鼠的索引顺序
order2存储进入下一轮老鼠的索引顺序

如果当前有groups个组，那么会有groups个老鼠进入下一轮，则没有进入下一轮的排名都为groups+1
如果只有一个组，那么最大的那个排名即为1。

(这段摘自http://www.cnblogs.com/chenxiwenruo/p/6519890.html)

这道题还是很有代表性的,但是还是可以把这道题从难到易的进行分解:

1,从头到尾遍历数组A,找到A的最大值---->太简单了

2,数组A从头到尾3个一组,找到每组内的最大值,并存到数组B中---->还行,不算难,就是多了一层对组的遍历,然后在每组内对"组员"遍历

3,数组A从头到尾3个一组,找到每组内的最大值,并还存到数组A中, ----->哈哈,这个也不难

4,重复3,直到数组A中只剩一个元素--->再来一层循环, 恩,稍加控制条件,不难

代码如下:

/*************************************************************************> File Name: 1056.c> Author: YueBo> Mail: [e-mail]> Created Time: Tue 23 May 2017 08:56:54 PM CST ************************************************************************/#include <stdio.h>#include <stdlib.h>int main(){    int weight[1024];    int programmer[1024];    int rank[1024];    int np, ng, first_np;    int i, j, k;    int rem;    int max_tmp;    scanf("%d%d", &np, &ng);    for (i = 0; i < np; i++)        scanf("%d", weight+i);    for (i = 0; i < np; i++)        scanf("%d", programmer+i);    first_np = np;    while (1)    {        rem = np % ng;        j = 0;        for (k = 0; k < np; k++)        {            max_tmp = -1;            for (i = ng*k; i < ng*k+ng && i < np; i++)            {                if (weight[programmer[i]] > max_tmp)                    max_tmp = weight[programmer[i]];            }            for (i = ng*k; i < ng*k+ng && i < np; i++)            {                if (weight[programmer[i]] == max_tmp)                {                    programmer[j] = programmer[i];                    j++;                }                else                    rank[programmer[i]] = np / ng + (rem==0 ? 0 : 1) + 1;            }        }        if (np == 1)        {            rank[programmer[0]] = 1;             break;        }        np = np / ng + (rem==0 ? 0 : 1);    }    for (i = 0; i < first_np; i++)    {        printf("%d", rank[i]);        printf(i==first_np-1 ? "":" ");    }    printf("\n");        return 0;}