1056 Mice and Rice (25point(s)) - C语言 PAT 甲级

1056 Mice and Rice (25point(s))

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 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

题目大意：

• 输入 np 只老鼠的体重，其序号从 0 递增，
• 第二行输入老鼠序号，即老鼠的出场顺序，每 ng 只老鼠为一组，每组中体重最大的老鼠晋级
• 按照老鼠的序号输出每只老鼠的排名

设计思路：

队列模拟晋级

• 队列存储老鼠出场的序号
• 每 ng 只老鼠中选择一只晋级，直到队列中剩余一只老鼠结束
  • 每一轮未晋级的老鼠排名等于，本轮次中晋级老鼠的数量加 1
  • 即等于本轮次中分组的数量加 1

编译器：C (gcc)

#include 

int main(void)
{
        int np, ng;
        int weights[1001] = {0}, ranks[1001] = {0}, queue[1001] = {0};
        int rank, max, count, index;
        int i;

        scanf("%d %d", &np, &ng);
        for (i = 0; i < np; i++) {
                scanf("%d", &weights[i]);
        }
        for (i = 0; i < np; i++) {
                scanf("%d", &queue[i]);
        }

        count = np;
        while (count > 1) {
                rank = (count % ng) ? (count / ng + 2): (count / ng + 1);
                index = 0;
                max = queue[0];
                for (i = 1; i < count; i++) {
                        if (i % ng == 0) {
                                queue[index] = max;
                                index++;
                                max = queue[i];
                        } else {
                                if (weights[queue[i]] > weights[max]) {
                                        ranks[max] = rank;
                                        max = queue[i];
                                } else {
                                        ranks[queue[i]] = rank;
                                }
                        }
                }
                queue[index] = max;
                index++;
                count = index;
        }
        ranks[queue[0]] = 1;

        printf("%d", ranks[0]);
        for (i = 1; i < np; i++) {
                printf(" %d", ranks[i]);
        }

        return 0;
}