pat 甲级 1056. Mice and Rice (25)

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 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

题意：模拟，设最多x个人一组，每一轮游戏先按照游玩者玩游戏的顺序分组，每一组产生一名分数最高者，其余的人止步这一轮比赛并获得(下一轮游玩人数)+1的名次，直至决定出第一名为止。
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<queue>
using namespace std;
#define INF 0x3f3f3f
#define N_MAX 100000+5
int n, group;
int weight[N_MAX],Rank[N_MAX];
vector<int>permute;
int main() {
    while (cin >> n >> group) {
        for (int i = 0; i < n; i++)scanf("%d", &weight[i]);
        for (int i = 0; i < n; i++) { int a; scanf("%d", &a); permute.push_back(a); }
        vector<int>vec = permute, vec2; int N = vec.size();
        while (N > group) {
            int rank = (N%group == 0) ? N / group + 1 : N / group + 2;
            int i = 0, tmpi = group;
            while (1) {
                int max_wei = 0, id;
                for (; i < tmpi; i++) {
                    if (max_wei < weight[vec[i]]) {
                        max_wei = weight[vec[i]];
                        id = vec[i];
                    }
                }
                vec2.push_back(id); //没落选的    
                for (int k = tmpi - group; k < tmpi; k++)//落选的确定排名
                    if (vec[k] != id)Rank[vec[k]] = rank;
                if (tmpi == N)break;
                tmpi += group;
                if (tmpi > N)tmpi = N;
            }
            vec = vec2;
            N = vec.size();
            vec2.clear();
        }
        if (N > 0) {//最后一组
            int max_wei = 0,id;
            for (int i = 0; i < vec.size();i++) {
                if (max_wei < weight[vec[i]]) {
                    max_wei = weight[vec[i]];
                    id = vec[i];
                }
            }
            for (int i = 0; i < vec.size();i++) {
                if (vec[i] != id)Rank[vec[i]] = 2;
                else Rank[vec[i]] = 1;
            }
        }

        for (int i = 0; i < n; i++)
            printf("%d%c", Rank[i], i + 1 == n ? '
' : ' ');
    }

    return 0;
}