1056 Mice and Rice (25 分)

1056 Mice and Rice (25 分)

 

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​​ (≤), 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​​ (,) 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 (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

开始眼睛一蒙，确实没看懂题意，纠结了半天。

#include <bits/stdc++.h>
using namespace std;
int n, m;
struct Node
{
    int val, pai, pos, pre;
    friend bool operator < (const Node &a,const Node &b){
        return a.pre < b.pre;
    }
}node[1005];
deque<Node> dq;
vector<Node> v;
int an[1005];

int main(){
    cin >> n >> m;
    for(int i = 0; i < n; i++){
        cin >> an[i];
    }
    int x;
    for(int i = 0; i < n ; i++){
        cin >> x;
        node[i].val = an[x];
        node[i].pos = i;
        node[i].pre = x;
        dq.push_back(node[i]);
    }
    while(dq.size() != 1){
        int len = dq.size();
        int group = len/m + (len%m == 0?0:1) + 1;
        for(int i = 0; i < n&&i < len; i=i+m){
            int ma = -1, id = -1;
            v.clear();
            for(int j = i; j < n&&j < len&&j < i+m; j++){
                if(ma < dq.front().val){
                    ma = dq.front().val;
                    id = dq.front().pos;
                }
                v.push_back(dq.front());
                dq.pop_front();
            }
            for(int j = 0; j < v.size(); j++){
                if(v[j].pos == id){
                    dq.push_back(v[j]);
                }else{
                    node[v[j].pos].pai = group;
                }
            }
        }
    }
    if(dq.size()==1)
        node[dq.front().pos].pai = 1;
    sort(node, node+n);
    for(int i = 0 ; i < n; i++)
        printf("%d%c", node[i].pai, i == n-1?'
':' ');
    return 0;
}