1056. Mice and Rice (25)

题目连接：https://www.patest.cn/contests/pat-a-practise/1056

原题如下：

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

又被虐了~
参考了http://blog.csdn.net/yangsongtao1991/article/details/43417363
首先得想到当前的rank等于晋级的个数+1；
其次每ng个进行比较的技巧；
最后，两个vector变量之间的赋值=；
不用struck即可，因为序号之间对应。

#include<stdio.h>
#include<vector>
using namespace std;
int main()
{
    vector<int>weight,order,next,rank;
    int np,ng;
    scanf("%d %d",&np,&ng);
    rank.resize(np);
    int v;
    for (int i=0;i<np;i++)
    {
        scanf("%d",&v);
        weight.push_back(v);
    }

    for (int i=0;i<np;i++)
    {
        scanf("%d",&v);
        order.push_back(v);
    }

    int curRank=0;
    int Maxw=-1,index;
    while(order.size()!=1)
    {
        int n=0;
        curRank=order.size()/ng+1;              //当前的名次是晋级人数+1；
        if ((order.size()%ng)!=0)curRank++;
        vector<int>().swap(next);
        while(n<order.size())
        {
            Maxw=-1;
            for (int i=0;i<ng && n<order.size();i++,n++)  //i<ng保证了每ng比较
        {
            rank[order[n]]=curRank;
            if (weight[order[n]]>Maxw)
            {
                Maxw=weight[order[n]];
                index=order[n];
            }
        }
        next.push_back(index);  //index是最大的
        }
        order=next;
    }
    rank[order[0]]=1;
    for (int i=0;i<np;i++)
    {
        if (i==0)printf("%d",rank[i]);
        else printf(" %d",rank[i]);
    }
    return 0;
}