1056 Mice and Rice（25 分）
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 NG 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: 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 Wi is 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 3
Sample Output:
5 5 5 2 5 5 5 3 1 3 5
一，正确解法
#include<cstdio>
#include<queue>
using namespace std;
//到9:00
struct mice_and_rice {
int weight;
int rank;
}mice[1010];
queue<int> que;
int main() {
int Np = 0;
int Ng = 0;
int num = 0;
int order = 0;
scanf("%d %d", &Np, &Ng);
for (int i = 0; i < Np; i++) {
scanf("%d", &mice[i].weight);
}
for (int i = 0; i < Np; i++) {
scanf("%d", &order);
que.push(order);
}
int temp = Np;
int group = 0;
while (temp != 1) {
if (temp%Ng == 0)group = temp / Ng;
else group = temp / Ng + 1;
for (int i = 0; i < group; i++) {
int max_mark = que.front();
for (int j = 0; j < Ng; j++) {
if (i*Ng + j >= temp)break;
int front = que.front();
if (mice[front].weight > mice[max_mark].weight) {
max_mark = front;
}
mice[front].rank = group + 1;
que.pop();
}
que.push(max_mark);
}
temp = group;
}
mice[que.front()].rank = 1;
for (int i = 0; i < Np; i++) {
printf("%d", mice[i].rank);
if (i != Np - 1) {
printf(" ");
}
}
return 0;
}二，总结
当初的困难是如何在队列中对那一个单位的三个体重进行比较，使得既能够把那三个体重分出高下（需要暂时保存它们的值），又能够比较完之后需要把那三个值pop(与暂时保存它们的值矛盾？)，同时又有如果一组中有两个甚至三个相等的值怎么办（事实上题干给出了这些值各不相等，未认真读题）。
解决方法是不必把那三个体重比出高下，只需要把它们归入那一轮，那一轮的排名统统有统一的排名即可。
三，
注：队列只能取和更改队首的元素，不能取和更改队尾的元素。