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
題意： 給你NP個老鼠以及他們的重量W，每NG個老鼠分一個組，不夠NG個數的單獨算一個組，比他們每個組的最大值，最大值進入下一輪的比較，同組其餘老鼠皆爲淘汰，並與其他組同時被淘汰的老鼠排名一致，求所有老鼠的排名。
思路：看着好像排個序就完事了，其實很麻煩，硬生生搞了一個小時，寫個博客記錄一下這個噁心的題，，，一開始沒想好要怎麼處理最後的那個排名，因爲它是每一層的失敗者排名相同，相當於，如果這一輪有x個人進入下一輪，那麼這些淘汰的人的排名就是x+1，剛開始把它想複雜了， 鄰接表都用上了，，服氣。。其他就沒什麼了，大體使用vector+結構體實現。
代碼：
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<iostream>
using namespace std;
#define inf 0x3f3f3f3f
const int N = 1005;
struct node
{
int grade,id;
}peo[N];
int ans[N];
vector<node> a,b,x;
bool cmp(node x,node y)
{
return x.grade>y.grade;
}
int main()
{
int n,k,xx;
cin>>n>>k;
for(int i=0;i<n;i++)
cin>>peo[i].grade,peo[i].id=i;
for(int i=1;i<=n;i++)
cin>>xx,a.push_back(peo[xx]);
while(a.size()>1) //a代表所有這一輪的人
{
x.clear();
for(int i=0;i<a.size();i+=k)
{
b.clear();
int sa=a.size();
for(int j=i;j<min(sa,i+k);j++)
b.push_back(a[j]); //b裏面存儲一組人
sort(b.begin(),b.end(),cmp);
x.push_back(b[0]); //x代表所有進了下一輪的人
}
for(int i=0;i<a.size();i++) //記錄排名
ans[a[i].id]=x.size()+1;
a=x; //進行迭代
}
ans[a[0].id]=1;
for(int i=0;i<n;i++)
{
if(i) cout<<" ";
cout<<ans[i];
}
return 0;
}