1056. Mice and Rice (25)

题目连接：https://www.patest.cn/contests/pat-a-practise/1056

原题如下：

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 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

又被虐了~
参考了http://blog.youkuaiyun.com/yangsongtao1991/article/details/43417363
首先得想到当前的rank等于晋级的个数+1；
其次每ng个进行比较的技巧；
最后，两个vector变量之间的赋值=；
不用struck即可，因为序号之间对应。

 1 #include<stdio.h>
 2 #include<vector>
 3 using namespace std;
 4 int main()
 5 {
 6     vector<int>weight,order,next,rank;
 7     int np,ng;
 8     scanf("%d %d",&np,&ng);
 9     rank.resize(np);
10     int v;
11     for (int i=0;i<np;i++)
12     {
13         scanf("%d",&v);
14         weight.push_back(v);
15     }
16 
17     for (int i=0;i<np;i++)
18     {
19         scanf("%d",&v);
20         order.push_back(v);
21     }
22 
23     int curRank=0;
24     int Maxw=-1,index;
25     while(order.size()!=1)
26     {
27         int n=0;
28         curRank=order.size()/ng+1;              //当前的名次是晋级人数+1；
29         if ((order.size()%ng)!=0)curRank++;
30         vector<int>().swap(next);
31         while(n<order.size())
32         {
33             Maxw=-1;
34             for (int i=0;i<ng && n<order.size();i++,n++)  //i<ng保证了每ng比较
35         {
36             rank[order[n]]=curRank;
37             if (weight[order[n]]>Maxw)
38             {
39                 Maxw=weight[order[n]];
40                 index=order[n];
41             }
42         }
43         next.push_back(index);  //index是最大的
44         }
45         order=next;
46     }
47     rank[order[0]]=1;
48     for (int i=0;i<np;i++)
49     {
50         if (i==0)printf("%d",rank[i]);
51         else printf(" %d",rank[i]);
52     }
53     return 0;
54 }

 

转载于:https://www.cnblogs.com/wuxiaotianC/p/6426746.html