URAL 1614. National Project “Trams” 构造

1614. National Project “Trams”

Time limit: 0.5 second

Memory limit: 64 MB

President has declared the development of tram service a priority national project. As a part of this project, Yekaterinburg will receive enough budget funds to carry out a complete reconstruction of
the city's tram network.

There are 2n tram stops in Yekaterinburg. In the course of reconstruction, the stops will be left in their places and no new stops will be built, but new tram railways will be laid so that
it will be possible to go by tram from every tram stop to any other tram stop without any intermediate stops.

Having studied messages at the tram forum, the city authorities found out that citizens would be satisfied with the reconstruction only if for every pair of tram stops there would be a tram route connecting
these stops without any intermediate stops. It is clear that the network of n(2n − 1) routes consisting of only two stops each satisfies this requirement. However, Yekaterinburg Mayor wants exactly n routes to be left after the reconstruction,
and each tram must make exactly 2n stops (including the final ones) on its route. Trams must go along these routes in both directions. Suggest a plan of reconstruction that will satisfy both citizens and Mayor.

Input

The only input line contains the integer n, 1 ≤ n ≤ 100.

Output

Output n lines describing tram routes. Each route is a sequence of integers in the range from 1 to 2n separated by a space. A route may go through a stop several times. If the problem
has several solutions, you may output any of them. If there is no solution, output one line containing the number −1.

Sample

input	output
3	1 6 2 1 3 4 2 3 6 5 4 6 5 1 4 2 5 3

Problem Author: Alexander Ipatov (idea — Magaz Asanov)

Problem Source: The 12th Urals Collegiate Programing Championship, March 29, 2008

Tags: tricky problem (

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)

题意：

输出n行，每行2*n个数。要求这2*n个数字能 两两相邻。

做法：

因为相邻的一共有n*(n-1)/2 对， 而把2*n个数字 两两互连，也是 n*(n-1)/2 对，所以每两个数之间是不能重复相邻的。

所以两侧 肯定是 123456 ，各个数字都出现一遍。

构造的方法是看别人的，首尾互取，然后把数列往后推1格，再首尾互取。

以n=3为例：

123456，

首尾互取 162534 为第一行

234561

首尾互取 213645 为第二行

345612

首尾互取 324156 为第三行

#include <cstdio>  
#include <cstring>  
#include <iostream>  
using namespace std;  
  
int N, M;  
  
int main() {  
    int i, j, k;  
    cin>>N;  
    M = N*2;  
    for (i = 0; i < N; ++i) {  
        int now = i;
        for (j = 0; j < N; ++j) {  
            printf("%d %d",(j+now)%M+1,(M-1-j+now)%M+1);
			if(j!=N-1)
				printf(" ");
        }  
        puts("");  
    }   
    return 0;  
}