usaco-4.4-shutter-passed

此题取巧了，Usaco在这题上并没有指明不可以用分析法，而且dfs肯定TLE，所以我们取巧。

先观察样例数据，如果把还没移动的那一步也算上，那么空格的位置为

4 3 5 6 4 2 1 3 5 7 6 4 2 3 5 4 (n=3,样例)

5 4 6 7 5 3 2 4 6 8 9 7 5 3 1 2 4 6 8 7 5 3 4 6 5 (n=4)

我们凭借极其敏锐的眼光发现这组序列为

４35 642 1357 642 35 4 (n=3,样例)

5 46 753 2468 97531 2468 753 46 5 (n=4)

即长度为1,2,3,4,...,n,n+1,n,...,4,3,2,1这样的2n+1组等差序列

我们讨论第1~n+1组序列，这些序列满足

　　*公差的绝对值为2

　　*奇数组为降序列，偶数组为升序列

　　*对于第i组(1<=i<=n+1),若为奇数组则首项为n+i，偶数组则首项为n-i+2

对于第n+2~2n+1组，可以由对称性求出。

输出时从第二组开始即可。

把规律总结成这样后，代码应该很好写了吧。

/*
ID: qq104801
LANG: C++
TASK: shuttle
*/

#include <iostream>
#include <fstream>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>

using namespace std;

void test()
{   
    freopen("shuttle.in","r",stdin);  
    freopen("shuttle.out","w",stdout);  
    int n,p,s=0,k;
    cin>>n;
    p=n+1;
    for(int i=2;i<=2*n+1;i++)
    {
        k=(i<=n+1)?i:2*n+2-i;
        p+=(k&1)^(i>n+1)?1:-1;
        s++;
        cout<<p<<((s%20==0||i==2*n+1) ? '\n' : ' ');
        for(int j=2;j<=k;j++)
            p+=(k&1)?-2:2,s++,
            cout<<p<<((s%20)?' ':'\n');
    } 
}

int main () 
{        
    test();        
    return 0;
}

test data:

USACO Training
Grader Results     
14 users online
BRA/1 CHN/4 IDN/1 IND/1 KOR/1 PER/1 USA/4 VNM/1

USER: cn tom [qq104801]
TASK: shuttle
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.008 secs, 3372 KB]
   Test 2: TEST OK [0.011 secs, 3372 KB]
   Test 3: TEST OK [0.011 secs, 3372 KB]
   Test 4: TEST OK [0.008 secs, 3372 KB]
   Test 5: TEST OK [0.005 secs, 3372 KB]
   Test 6: TEST OK [0.008 secs, 3372 KB]
   Test 7: TEST OK [0.008 secs, 3372 KB]
   Test 8: TEST OK [0.011 secs, 3372 KB]
   Test 9: TEST OK [0.008 secs, 3372 KB]
   Test 10: TEST OK [0.016 secs, 3372 KB]

All tests OK.

YOUR PROGRAM ('shuttle') WORKED FIRST TIME! That's fantastic -- and a rare thing. Please accept these special automated congratulations.

Here are the test data inputs:

------- test 1 ----
1
------- test 2 ----
3
------- test 3 ----
4
------- test 4 ----
5
------- test 5 ----
7
------- test 6 ----
8
------- test 7 ----
9
------- test 8 ----
10
------- test 9 ----
11
------- test 10 ----
12

Keep up the good work!
Thanks for your submission!