目录
题面
Square Ice
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 4526 | Accepted: 1759 |
Description
Square Ice is a two-dimensional arrangement of water molecules H2O, with oxygen at the vertices of a square lattice and one hydrogen atom between each pair of adjacent oxygen atoms. The hydrogen atoms must stick out on the left and right sides but are not allowed to stick out the top or bottom. One 5 x 5 example is shown below.
Note that each hydrogen atom is attached to exactly one of its neighboring oxygen atoms and each oxygen atom is attached to two of its neighboring hydrogen atoms. (Recall that one water molecule is a unit of one O linked to two H's.)
It turns out we can encode a square ice pattern with what is known as an alternating sign matrix (ASM): horizontal molecules are encoded as 1, vertical molecules are encoded as -1 and all other molecules are encoded as 0. So, the above pattern would be encoded as:
An ASM is a square matrix with entries 0, 1 and -1, where the sum of each row and column is 1 and the non-zero entries in each row and in each column must alternate in sign. (It turns out there is a one-to-one correspondence between ASM's and square ice patterns!)
Your job is to display the square ice pattern, in the same format as the example above, for a given ASM. Use dashes (-) for horizontal attachments and vertical bars (|) for vertical attachments. The pattern should be surrounded with a border of asterisks (*), be left justified and there should be exactly one character between neighboring hydrogen atoms (H) and oxygen atoms (O): either a space, a dash or a vertical bar.
Input
Input consists of multiple cases. Each case consists of a positive integer m (<= 11) on a line followed by m lines giving the entries of an ASM. Each line gives a row of the ASM with entries separated by a single space. The end of input is indicated by a line containing m = 0.
Output
For each case, print the case number (starting from 1), in the format shown in the Sample Output, followed by a blank line, followed by the corresponding square ice pattern in the format described above. Separate the output of different cases by a blank line.
Sample Input
2
0 1
1 0
4
0 1 0 0
1 -1 0 1
0 0 1 0
0 1 0 0
0 Sample Output
Case 1:
***********
*H-O H-O-H*
* | *
* H H *
* | *
*H-O-H O-H*
***********
Case 2:
*******************
*H-O H-O-H O-H O-H*
* | | | *
* H H H H *
* | *
*H-O-H O H-O H-O-H*
* | | *
* H H H H *
* | | *
*H-O H-O H-O-H O-H*
* | *
* H H H H *
* | | | *
*H-O H-O-H O-H O-H*
******************* Source
East Central North America 2001
思路
没怎么看懂题意。但是知道了这么几个点：
1. 竖直的水分子不会出现在第一行和最后一行。
2. 给出的矩阵1,-1是交替出现的。
从样例输出观察出了输出行数，输出列数，H原子，O原子的规律。
不包括边界星号的话，行数是4*m-3，列数是4*m+1.
思路
先把星号，氧原子，氢原子放好，其余空格填充。
另外开一个数组该位置的原子已经连了几个键。
首先根据输入把水平水分子和竖直水分子的键连了，并且更新“连键”数组。
H原子分成两种类型。
一种是其键只能水平连的（横纵坐标都是模4余1），一种是其键只能竖直连的（横纵坐标都是模4余3）。
先只考虑水平型，我们连向与它相邻的O原子（必须空余键值为0，否则可能可能两个水平型H原子连到一个O原子变成一个水平水分子）。
按理来说，处理完完水平型的H原子，应该没有O原子剩余键数是2（否则它就只能通过两个竖直型H原子来连，那样就构成了竖直型水分子）。
然后考虑竖直型H原子，显然只能连到相邻的已经连了一个H原子的O原子上。
因为没怎么看懂题意，所以是连蒙带猜的，为了好检验我是否猜错了。我的代码中见了抛出异常。
如果我猜错了，那么就会抛出异常，评测状态是RE.
结果WA了好几次——因为Case,空行的格式
AC代码
#include <iostream>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
const int kMaxSquareSize = 11;
char board[4*kMaxSquareSize-1][4*(kMaxSquareSize+1)];
int link_cnt[4*(kMaxSquareSize)+3][4*(kMaxSquareSize+1)];
int in[kMaxSquareSize][kMaxSquareSize];
inline bool is_border(int h,int w,int i,int j) {
return i == 0 || i+1 == h || j == 0 || j+1 == w;
}
inline bool is_oxygen(int i,int j) {
return i%4 == 1 && j%4 == 3;
}
inline bool is_hydrogen(int i,int j) {
return (i%4 == 3 && j%4 == 3) || (i%4 == 1 && j%4 == 1);
}
int main()
{
int m;
memset(board,0,sizeof(board));
int id = 0;
while (scanf("%d",&m) && m) {
if (id)
printf("\n");
printf("Case %d:\n\n",++id);
memset(link_cnt,false,sizeof(link_cnt));
int h = m*4-1, w = m*4+3;
for (int i = 0;i < h; ++i)
for (int j = 0;j < w; ++j) {
if (is_border(h,w,i,j))
board[i][j] = '*';
else if (is_oxygen(i,j))
board[i][j] = 'O';
else if (is_hydrogen(i,j))
board[i][j] = 'H';
else
board[i][j] = ' ';
}
for (int i = 0;i < m; ++i)
for (int j = 0;j < m; ++j) {
scanf("%d",in[i]+j);
if (in[i][j] == 1) {
board[4*i+1][4*j+2] = '-'; link_cnt[4*i+1][4*j+1] = 1;
board[4*i+1][4*j+4] = '-'; link_cnt[4*i+1][4*j+5] = 1;
link_cnt[4*i+1][4*j+3] = 2;
} else if (in[i][j] == -1) {
board[4*i][4*j+3] = '|'; link_cnt[4*i-1][4*j+3] = 1;
board[4*i+2][4*j+3] = '|'; link_cnt[4*i+3][4*j+3] = 1;
link_cnt[4*i+1][4*j+3] = 2;
}
}
// 1型氢原子，水平短线连接
for (int i = 1;i < h; i += 4)
for (int j = 1;j < w; j += 4) {
if (link_cnt[i][j]) continue;
if ((j > 3) && (link_cnt[i][j-2] == 0)) {
board[i][j-1] = '-';
++link_cnt[i][j-2];
link_cnt[i][j] = 1;
} else if ((j + 2 < w) && (link_cnt[i][j+2] == 0)) {
board[i][j+1] = '-';
++link_cnt[i][j+2];
link_cnt[i][j] = 1;
} else {
//cout<<i<<","<<j<<endl;
throw 5.0;
}
}
// 2型 竖直线连接
for (int i = 3;i < h; i += 4)
for (int j = 3;j < w; j += 4) {
if (link_cnt[i][j]) continue;
if (link_cnt[i-2][j] == 1) {
board[i-1][j] = '|';
++link_cnt[i-2][j];
link_cnt[i][j] = 1;
} else if (link_cnt[i+2][j] == 1) {
board[i+1][j] = '|';
++link_cnt[i+2][j];
link_cnt[i][j]= 1;
} else {
//cout<<i<<","<<j<<endl;
throw "wrong";
}
}
for (int i = 0;i < h; ++i) {
board[i][w] = 0;
puts(board[i]);
}
}
return 0;
}
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