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 NGwinners 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 Wiis 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 3Sample Output:
5 5 5 2 5 5 5 3 1 3 5
备注:题目略长,读完才知道就是每一轮分组比较大小,通过多轮选出优胜者,最后输出排名。解题思路是:每一轮结束就可以确定loser rats的rank(winner rats的数量+1),然后继续下一轮,直到最后找到final winner。刚开始有2个case一直超时,后来才发现可以把每轮winner rats的index存在数组里,这样就可以大大减少下一轮的扫描次数。
#include<stdio.h> #include<string.h> const int NUM = 1001; const int LOSER = -1; const int WINNER = -2; int main() { int Np,Ng; int w[NUM],order[NUM]; int rank[NUM]; scanf("%d %d",&Np,&Ng); for(int i=0;i<Np;i++) scanf("%d",&w[i]); for(int i=0;i<Np;i++) scanf("%d",&order[i]); int count = 0, count_update = 0, k = 0; int max = -1,maxIndex = -1; int last_winner = -1; int r,n,n_rats = Np; memset(rank,LOSER,sizeof(rank)); while(true) { r = n_rats/Ng + ((n_rats%Ng==0)?0:1); n_rats = r; n = Np-count_update; r++; k = 0; for(int i=0;i<n;i++) { int index = order[i]; if(count<Ng) { count++; if(w[index]>max) { max = w[index]; maxIndex = index; } } if((count==Ng || i==n-1) && max!=-1) { rank[maxIndex] = WINNER; order[k++] = maxIndex; // store the winner's index, so in next round the scan length can be reduced dramatically max = -1; maxIndex = -1; count = 0; } } for(int i=0;i<Np;i++) { if(rank[i]==LOSER) // loser rats, whose ranking are to be set { count_update++; rank[i] = r; } else if(rank[i]==WINNER) // winner rats, who need to be ranked later { rank[i] = LOSER; last_winner = i; } } if(count_update == Np-1) break; } if(last_winner!=-1) rank[last_winner]=1; for(int i=0;i<Np;i++) { if(i==Np-1) printf("%d",rank[i]); else printf("%d ",rank[i]); } return 0; }