1056 Mice and Rice （PAT甲级）

这道题我觉得柳婼的解法有点把题复杂化了，其实简单模拟就可以完成。cnt是当前的group数，会选出cnt个进入下一轮比赛，所以当前被淘汰的排名为cnt + 1。

又想了下，觉得柳婼这里queue用得还是挺巧妙的，根据她的思路把她的解法稍微简化了下，这是我的版本。

#include 
#include 
#include 

int N, g, maxx, pivot, cnt, t, sz;
std::vector w, rank;
std::queue perm;

int main(){
    scanf("%d %d", &N, &g);
    w.resize(N);
    rank.resize(N);
    for(int i = 0; i < N; ++i){
        scanf("%d", &w[i]);
    }
    for(int i = 0; i < N; ++i){
        scanf("%d", &t);
        perm.push(t);
    }
    while(!perm.empty()){
        sz = perm.size();
        if(sz == 1){
            rank[perm.front()] = 1;
            break;
        }
        cnt = perm.size() % g ? perm.size() / g + 1 : perm.size() / g;
        maxx = -1;
        for(int i = 0; i < sz; ++i){
            if(w[perm.front()] > maxx){
                pivot = perm.front();
                maxx = w[perm.front()];
            }
            rank[perm.front()] = cnt + 1;
            perm.pop();
            if(i == sz - 1 || i % g == g - 1){
                perm.push(pivot);
                maxx = -1;
            }
        }
    }
    for(int i = 0; i < N; ++i){
        printf("%d%s", rank[i], i == N - 1 ? "\n" : " ");
    }
    return 0;
}

下面是我原来的版本：

#include 
#include 

int N, g, maxx, pivot, cnt, tmp;
std::vector w, perm, rank, curr;

int main(){
    scanf("%d %d", &N, &g);
    w.resize(N);
    perm.resize(N);
    rank.resize(N);
    for(int i = 0; i < N; ++i){
        scanf("%d", &w[i]);
    }
    for(int i = 0; i < N; ++i){
        scanf("%d", &perm[i]);
    }
    while(perm.size() != 1){
        curr.clear();
        cnt = perm.size() % g ? perm.size() / g + 1 : perm.size() / g;
        for(int i = 0; i < cnt; ++i){
            maxx = -1;
            tmp = perm.size() % g ? perm.size() % g : g;
            for(int j = 0; j < (i == cnt - 1 ? tmp : g); ++j){
                if(w[perm[i * g + j]] > maxx){
                    pivot = j;
                    maxx = w[perm[i * g + j]];
                }
            }
            curr.push_back(perm[i * g + pivot]);
            for(int j = 0; j < (i == cnt - 1 ? tmp : g); ++j){
                if(j != pivot){
                    rank[perm[i * g + j]] = cnt + 1;
                }
            }
        }
        perm = curr;
    }
    rank[perm[0]] = 1;
    for(int i = 0; i < N; ++i){
        printf("%d%s", rank[i], i == N - 1 ? "\n" : " ");
    }
    return 0;
}

题目如下:

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 NP​ programmers. Then every NG​ 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 NG​ 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: NP​ and NG​ (≤1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than NG​ 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 NP​ distinct non-negative numbers Wi​ (i=0,⋯,NP​−1) where each Wi​ is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,⋯,NP​−1 (assume that the programmers are numbered from 0 to NP​−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