2017ACM-ICPC亚洲区域赛北京站 F.Secret Poems(模拟)

描述

The Yongzheng Emperor (13 December 1678 – 8 October 1735), was the
fifth emperor of the Qing dynasty of China. He was a very hard-working
ruler. He cracked down on corruption and his reign was known for being
despotic, efficient, and vigorous.

Yongzheng couldn’t tolerate people saying bad words about Qing or him.
So he started a movement called “words prison”. “Words prison” means
literary inquisition. In the famous Zhuang Tinglong Case, more than 70
people were executed in three years because of the publication of an
unauthorized history of the Ming dynasty.

In short, people under Yongzheng’s reign should be very careful if
they wanted to write something. So some poets wrote poems in a very
odd way that only people in their friends circle could read. This kind
of poems were called secret poems.

A secret poem is a N×N matrix of characters which looks like random
and meaning nothing. But if you read the characters in a certain
order, you will understand it. The order is shown in figure 1 below:

Following the order indicated by arrows, you can get
“THISISAVERYGOODPOEMITHINK”, and that can mean something.

But after some time, poets found out that some Yongzheng’s secret
agent called “Mr. blood dripping” could read this kind of poems too.
That was dangerous. So they introduced a new order of writing poems as
shown in figure 2. And they wanted to convert the old poems written in
old order as figure1 into the ones in new order. Please help them.

输入

There are no more than 10 test cases.

For each test case:

The first line is an integer N( 1 <= N <= 100), indicating that a poem
is a N×N matrix which consist of capital letters.

Then N lines follow, each line is an N letters string. These N lines
represent a poem in old order.

输出

For each test case, convert the poem in old order into a poem in new
order.

样例输入

5THSAD IIVOP SEOOH RGETI YMINK2ABCD4ABCDEFGHIJKLMNOP

样例输出

THISIPOEMSDNKIAOIHTVOGYREABDCABEIKHLFNPOCMJGD

思路

题意很简单，看那两张图就可以明白，先用第一张图的方式把图片中的序列读出来，再用类似蛇形填数的方法，把这一串序列填进去，然后输出填后的图就可以

代码

#include <bits/stdc++.h>using namespace std;#define mem(a,b) memset(a,b,sizeof(a))const int N=100+20;int n;char mp1[N][N];int mp2[N][N];string init(){    string s="";    s+=mp1[1][1];    int x=1,y=1,op=0;    while(!(x==n&&y==n))    {        if(x==1&&op==0)        {            if(y+1<=n) y++;            else x++;            op=1;        }        else if(y==1&&op==1)        {            if(x+1<=n) x++;            else y++;            op=0;        }        else        {            if(op==1)            {                if(x+1<=n&&y-1>=1)                {                    x++;                    y--;                    op=1;                }                else                {                    y++;                    op=0;                }            }            else if(op==0)            {                if(x-1>=1&&y+1<=n)                {                    x--;                    y++;                    op=0;                }                else                {                    x++;                    op=1;                }            }        }        s+=mp1[x][y];    }    return s;}void solve(){    mem(mp2,0);    int x=0,y=0,cnt;    cnt=mp2[x][y]=1;    while(cnt<n*n)    {        while(y+1<n&&!mp2[x][y+1])  mp2[x][++y]=++cnt;        while(x+1<n&&!mp2[x+1][y])  mp2[++x][y]=++cnt;        while(x-1>=0&&!mp2[x-1][y]) mp2[--x][y]=++cnt;        while(y-1>=0&&!mp2[x][y-1]) mp2[x][--y]=++cnt;    }}int main(){    while(~scanf("%d",&n))    {        for(int i=1; i<=n; i++)            scanf("%s",mp1[i]+1);        string s=init();        solve();        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                printf("%c",s[mp2[i][j]-1]);            }            puts("");        }    }    return 0;;}