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。 1 #include <bits/stdc++.h> 2 using namespace std; 3 int n, m; 4 struct Node 5 { 6 int val, pai, pos, pre; 7 friend bool operator < (const Node &a,const Node &b){ 8 return a.pre < b.pre; 9 } 10 }node[1005]; 11 deque<Node> dq; 12 vector<Node> v; 13 int an[1005]; 14 15 int main(){ 16 cin >> n >> m; 17 for(int i = 0; i < n; i++){ 18 cin >> an[i]; 19 } 20 int x; 21 for(int i = 0; i < n ; i++){ 22 cin >> x; 23 node[i].val = an[x]; 24 node[i].pos = i; 25 node[i].pre = x; 26 dq.push_back(node[i]); 27 } 28 while(dq.size() != 1){ 29 int len = dq.size(); 30 int group = len/m + (len%m == 0?0:1) + 1; 31 for(int i = 0; i < n&&i < len; i=i+m){ 32 int ma = -1, id = -1; 33 v.clear(); 34 for(int j = i; j < n&&j < len&&j < i+m; j++){ 35 if(ma < dq.front().val){ 36 ma = dq.front().val; 37 id = dq.front().pos; 38 } 39 v.push_back(dq.front()); 40 dq.pop_front(); 41 } 42 for(int j = 0; j < v.size(); j++){ 43 if(v[j].pos == id){ 44 dq.push_back(v[j]); 45 }else{ 46 node[v[j].pos].pai = group; 47 } 48 } 49 } 50 } 51 if(dq.size()==1) 52 node[dq.front().pos].pai = 1; 53 sort(node, node+n); 54 for(int i = 0 ; i < n; i++) 55 printf("%d%c", node[i].pai, i == n-1?'\n':' '); 56 return 0; 57 }

 

转载于:https://www.cnblogs.com/zllwxm123/p/11187268.html