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 NGprogrammers 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
核心思想
这道题目的难度在于你的算法及选择的容器，第一你要完全模拟分组比赛，第二你要给他们排名。这里选择数组作为存储容器，第一个数组存储他的体重，Ng存储他的分组数，
完整代码
#include<iostream>
#include<vector>
using namespace std;
int Np,Ng;
vector<int>players;
int main()
{
//freopen("C:\\Users\\Administrator\\Desktop\\test\\input.txt","r",stdin);
int i,j,k,l,score[1000],Rank[1000];
cin >> Np >> Ng;
for(i=0;i<Np;i++) cin >> score[i];
for(i=0;i<Np;i++){
cin >> j;
players.emplace_back(j);
}
for(int num = Np;num>1;){
vector<int>nextturn;
for(i=0;i<num;i+=Ng){
int jinji = i;
for(j = i+1;j<i+Ng&&j<num;j++){
if(score[players[j]]>score[players[jinji]]){
jinji = j;
}
}
nextturn.emplace_back(players[jinji]);
for(j=i;j<i+Ng&&j<num;j++){
if(j!=jinji){
Rank[players[j]] = num/Ng+(num%Ng?1:0)+1;
}
}
}
players = nextturn;
num = num/Ng+(num%Ng?1:0);
}
Rank[players[0]] = 1;
for(i=0;i<Np;i++){
if(i) cout << " ";
cout << Rank[i];
}
return 0;
}
方法2利用队列做
#include<cstdio>
#include<stack>
#include<queue>
using namespace std;
const int maxn = 1010;
struct mouse{ //老鼠
int weight;//质量
int R;//排名
}mouse[maxn];
int main()
{
int np,ng,order;
scanf("%d%d",&np,&ng);
for(int i =0;i<np;i++){
scanf("%d",&mouse[i].weight);
}
queue<int> q;//定义一个队列
for(int i =0;i<np;i++){
scanf("%d",&order);//题目给出的顺序
q.push(order); //按顺序把老鼠们的标号入队
}
int tmp = np,group; //temp为当前轮的比赛老鼠数，group为组数
while(q.size()!=1){
//计算group，即当前轮分为几组进行比赛
if(tmp%ng==0) group = tmp/ng;
else group = tmp/ng+1;
//枚举每一组，选出该组老鼠中质量最大的
for(int i =0;i<group;i++){
int k = q.front();//k存储该组质量最大的老鼠编号
for(int j =0;j<ng;j++){
//在最后一组老鼠数不足Ng起作用，退出循环
if(i*ng+j>=tmp) break;
int front = q.front();//队首老鼠编号
if(mouse[front].weight>mouse[k].weight){
k = front;
}
mouse[front].R = group+1;
q.pop();
}
q.push(k);//把胜利1的老鼠晋级
}
tmp = group; // group 只老鼠晋级，因此下轮棕老师为group
}
mouse[q.front()].R = 1;
for(int i = 0;i<np;i++){
printf("%d",mouse[i].R);
if(i<np-1) printf(" ");
}
return 0;
}
