PAT1056

题目链接:http://pat.zju.edu.cn/contests/pat-a-practise/1056

思路，每次选出符合条件的然后放在新的数组里，一直循环直到剩一个，然后更新排名，此轮losers排名=这轮winners个数+1.

#include<cstdio>
#include<vector>
using namespace std;

int main()
{
    int N,M;
    scanf("%d%d",&N,&M);
    vector<int> weight(N);
    vector<int> outcome(N,0);
    vector<int> order[1000];
    int length(0);
    for(int i=0; i<N; ++i)
        scanf("%d", &weight[i]);
    for(int i=0; i<N; ++i)
    {
        int d;
        scanf("%d", &d);
        order[length].push_back(d);
    }
    ++length;
    while(order[length-1].size() != 1)
    {
        int count =order[length-1].size();
        if(count % M == 0)
            count /= M;
        else
            count = count/M+1;
        for(int i=0; i<count; ++i)
        {
            int max(-1), index(-1);
            for(int j=i*M; j<i*M+M && j<order[length-1].size(); ++j)
            {
                if(weight[order[length-1][j]] > max)
                {
                    max = weight[order[length-1][j]];
                    index = order[length-1][j];
                }
            }
            order[length].push_back(index);
        }
        ++length;
    }
    for(int i=length-1; i>=0; --i)
    {
        if(order[i].size()==1)
            outcome[order[i][0]] = 1;
        else
        {
            int rank = order[i+1].size()+1;
            for(int j=0; j<order[i].size(); ++j)
            if(outcome[order[i][j]] == 0)
                outcome[order[i][j]] = rank;
        }
    }
    for(int i=0; i<N-1; ++i)
        printf("%d ", outcome[i]);
    printf("%d
", outcome[N-1]);
    return 0;
}