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
这道题好绕啊， 读不懂！！
第三行的数字不是第i位选手的比赛顺序，而是比赛选手的序号； 如上面的sample input的第三行， 6 0 8 第一组的三个人的序号是6 0 8， 而不是第一个人的比赛顺序是第六；
思路：建立一个结构保存选手老鼠的重量以及排名；
用queue来保存晋级选手的序号
如何确定比赛名次？ 用第一轮比赛来举例：第一轮有11人， 3人一组，则分为4组， 一定有四个人晋级，那么淘汰的人的名次就在这四人之后，即排名为5； 第二轮有4个人可以分成2组，
可以晋级2人，这一轮比赛淘汰的名次为3
如何确定每一轮的比赛次数？ 可以发现queue中的元素个数就是该轮比赛的人数， 每一轮的比赛次数与这一轮的分组数目相等，则比赛次数 rank=q.size()/g + (q.size()%g==0 ? 0 : 1)
加好后面加的数，是为了解决比赛人数有剩余的情况(g表示每一组的人数)
如何找到每一组比赛重量最大的老鼠？ 采用顺序遍历的方法，每次遍历弹出queue最前面的元素，即比赛选手的序号， 遍历每一组中的每一个选手，就能找出最终选手的序号， 并且在该过程中更新选手排名
把每一轮每一组比赛的优胜者的序号添加到queue<int> temp中，改队列中记录的就是下一轮比赛的选手序号， 一轮比赛完成后， 把temp的值复制到queue中
注意点：注意循环的中值条件，应该是q.size()!=1, 而非去，q.size（）!=0; 否则会陷入死循环
对于在循环中长度会变化的结构来说，不能用i<q.size() 来判断，这样会导致错误，应该在循环前取得q的长度，再做判断
1 #include<iostream> 2 #include<queue> 3 #include<algorithm> 4 using namespace std; 5 struct node{ int weight, rank;}; 6 7 int main(){ 8 int n, g, i, j, index, maxid; 9 cin>>n>>g; 10 queue<int> q; 11 vector<node> v(n); 12 for(i=0; i<n; i++) cin>>v[i].weight; 13 for(i=0; i<n; i++){ cin>>index; q.push(index);} 14 15 while(q.size()!=1){ 16 int rank = q.size()/g + (q.size()%g==0 ? 0 : 1); 17 queue<int> temp; 18 for(i=0; i<rank; i++){ 19 maxid=q.front(); 20 int len=q.size(); 21 for(j=0; j<g && j<len; j++){ 22 int idxx=q.front(); 23 if(v[maxid].weight < v[idxx].weight) maxid=idxx; 24 v[idxx].rank=rank+1; 25 q.pop(); 26 } 27 temp.push(maxid); 28 } 29 q=temp; 30 } 31 32 v[maxid].rank=1; 33 cout<<v[0].rank; 34 for(i=1; i<n; i++) cout<<" "<<v[i].rank; 35 return 0; 36 }