A1056 Mice and Rice (25分)【C语言】
原题链接
题目描述：
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.
输入格式：
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.
输出格式：
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.
输入样例：
11 3
25 18 0 46 37 3 19 22 57 56 10
6 0 8 7 10 5 9 1 4 2 3
输出样例：
5 5 5 2 5 5 5 3 1 3 5
实现代码：
#include <cstdio>
#include <queue>
using namespace std;
const int maxn = 1010;
queue<int> q; //队列存储比赛老鼠编号顺序 struct mouse{int weigh; //重量 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].weigh);}for(int i=0; i<np; ++i){scanf("%d", &order); //题目中所给分组顺序q.push(order); //把题目中所给最初分组顺序输入队列 }int temp=np, group; //temp为当前轮参赛老鼠数，group为当前轮的组数 while(q.size()!=1){ //只要队列中还有老鼠 if(temp%ng==0){group = temp / ng;}else{group = temp / ng + 1; //最后一组老鼠数不够一组数 }//枚举每一组，选出该组中质量最大的老鼠for(int i=0; i<group; ++i){int k = q.front(); //k存放当前组质量最大老鼠的编号for(int j=0; j<ng; ++j){if(i*ng+j>=temp) break; //最后一组老鼠数量不够，退出循环int front = q.front();if(mouse[front].weigh>mouse[k].weigh){k = front; //更新质量最大的老鼠 }mouse[front].R = group+1; //每次都把参与该轮的老鼠排名更新为group+1q.pop(); //参加完该轮比赛的老鼠弹出队列 }q.push(k); //将晋级的老鼠加入队列，参加下一轮比赛 } temp = group; //group只老鼠晋级，因此下一轮参加比赛的老鼠更新为group }mouse[q.front()].R = 1; //队列中只剩一只老鼠时，它就是第一for(int i=0; i<np; ++i){printf("%d", mouse[i].R);if(i<np-1) printf(" ");} return 0;
}