1056. Mice and Rice (25)
时间限制
30 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Mice and Riceis 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 NPprogrammers. 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 NGwinners 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: NPand NG(<= 1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than NGmice at the end of the player’s list, then all the mice left will be put into the last group. The second line contains NPdistinct non-negative numbers Wi(i=0,…NP-1) where each Wiis 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:11 325 18 0 46 37 3 19 22 57 56 106 0 8 7 10 5 9 1 4 2 3Sample Output:5 5 5 2 5 5 5 3 1 3 5
做这道题真是一口老血吐出来了 = =。一开始无法理解题意，都看不懂那个sample output是怎么来的，后来发现是题目没看仔细 = =
比如sample input中，第三行的第一个数字是6，意思是6号老鼠排第一位，而不是1号老鼠排第六位。。。所以说看清题目很重要千万不要自己YY啊。。
另外一个就是rank问题，这个rank不是按照轮次来的，，而是下一轮晋升数+1.
搞清了题意以后就简单了，代码如下：
#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <cmath>#include <queue>using namespace std;typedef struct player{int id;int weight;int rank;int order;}player;bool cmp1(player a,player b){return a.order<b.order;}bool cmp2(player a,player b){return a.weight<b.weight;}bool cmp3(player a,player b){return a.id<b.id;}int main(){int Np,Ng,i;cin>>Np>>Ng;vector<player> ori(Np);vector<player> judge;for(i=0;i<Np;i++){ori[i].id=i;cin>>ori[i].weight;}for(i=0;i<Np;i++){int temp;cin>>temp;ori[temp].order=i;}sort(ori.begin(),ori.end(),cmp1);queue<player> game;while(true){for(i=0;i<Np;i++){if(ori[i].rank==0)game.push(ori[i]);}if(game.size()==1){player temp=game.front();ori[temp.order].rank=1;break;}int count=ceil(game.size()*1.0/Ng)+1;while(game.size()){int j=Ng;judge.clear();while(j–&&game.size()>0){judge.push_back(game.front());game.pop();}int size=judge.size();if(size==1)continue;sort(judge.begin(),judge.end(),cmp2);for(int k=0;k<size-1;k++)ori[judge[k].order].rank=count;}count++;}sort(ori.begin(),ori.end(),cmp3);for(i=0;i<Np-1;i++){cout<<ori[i].rank<<" ";}cout<<ori[Np-1].rank;}
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最重要的是今天的心。
