第十四届华中科技大学程序设计竞赛 F:Sorting Trees

本文介绍了一个树排序问题,通过冒泡排序算法对校园内的树木按高度进行升序排列,但因程序员失误导致排序错误。文章提供了一种方法来确定排序过程中首次出现错误的位置。
链接: https://www.nowcoder.com/acm/contest/106/F
来源:牛客网

It’s universally acknowledged that there’re innumerable trees in the campus of HUST.

One day the tree manager of HUST asked you to sort N trees increasing by their heights a i. Because you were a warm-hearted and professional programmer, you immediately wrote the  Bubble Sort to finish the task. However, you were so tired that made a huge mistake:
for(int i=1;i<n;i++)
    for(int j=1;j<=n-i;j++)
         if(a[j+k]<a[j])swap(a[j+k],a[j]);
Usually k=1 is the normal  Bubble Sort. But you didn't type it correctly. When you now realize it, the trees have been replanted. In order to fix them you must find the first place that goes wrong, compared with the correctly sorted tree sequence.

If every trees are in their correct places then print -1. And you should know k can be equal to 1 which means you didn't make any mistake at all. Also you don't need to consider the part j+k beyond n.

输入描述:


The first line contains two integers N and K as described above.
Then the next line are N integers indicating the unsorted trees.

输出描述:

A integer in a single line, indicating the first place that goes wrong. Or -1 means no mistake.
#include <bits/stdc++.h>
 
using namespace std;
const int N = 1E5 + 7;
int a[N];
bool vis[N];
int b[N];
int main()
{
    int n, k;
    scanf("%d %d", &n, &k);
    for(int i = 1;i <= n;i ++) {
        scanf("%d", &a[i]);
    }
    if(k == 0) {
        for(int i = 1;i <= n;i ++) b[i]=a[i];
        sort(a+1,a+1+n);
        int pos=-1;
        for(int i = 1;i <= n;i ++) {
            if(a[i] != b[i]) {
                pos = i;
                break;
            }
        }
        return !printf("%d\n", pos);
    }
    vector<vector<pair<int,int> > >  v;
    for(int i = 1;i <= n;i ++) {
        if(vis[i]) continue;
        vector<pair<int,int> > tmp;
        vector<int> pos;
        for(int j = i;j <= n;j += k) {
            vis[j] = 1;
            pos.push_back(j);
            tmp.push_back(make_pair(a[j], j) );
        }
        if(!tmp.empty()) {
            sort(tmp.begin(), tmp.end());
            for(int j = 0;j < pos.size();j ++) {
                tmp[j].second = pos[j];
            }
            v.push_back(tmp);
        }
    }
    vector<int> b(n+1);
    for(int i = 0;i < v.size();i ++) {
        for(int j = 0;j < v[i].size();j ++) {
            b[v[i][j].second] = v[i][j].first;
        }
    }
    sort(a+1, a+1+n);
    int pos = -1;
    for(int i = 1;i <= n;i ++) {
        if(a[i] != b[i]) {
            pos = i;
            break;
        }
    }
    printf("%d\n", pos);
    return 0;
}

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