ZOJ 3810 A Volcanic Island（四色定理模板）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5342

A Volcanic Island

After a lot of multilateral negotiation and occasional armed conflicts, the
N countries decided to divide the square volcanic island equally. They partitioned the island into
N x N small equal-sized square chunks. Each country could get a connected region consists of exact
N chunks.

Two chunks A and B are called "connected" if they share an edge, or there exists another chunk C connected with both A and B. A group of chunks are called "connected region" if any two of these chunks are connected.

Every country want a unique region. It means the N regions should be different with each other. Two regions are considered as the same if and only if one can transform into the other by an isometry (a combination of rigid motions, including translation,
rotation and reflection).

In a nutshell, your task is to divide a square island with N x
N chunks into N connected regions with different shape. You also need to draw a map to color the regions of the map so that no two edge-adjacent regions have the same color. Most of the people in these countries believed that four different
colors are enough. So you can mark these regions with at most four colors, red, green, blue and yellow.

Input

There are multiple test cases. The first line of input contains an integer
T indicating the number of test cases. For each test case:

There is only an integer N (1 <= N <= 100).

Output

For each test case, output a valid map described above. If there is no solution, output "No solution!" instead. Please note that only four colors (‘R‘, ‘G‘, ‘B‘ and ‘Y‘) can be used to drawing the map.

Sample Input

2
2
5

Sample Output

No solution!
YYYGR
YGGGR
YGYYR
BYYYR
BBBBR

Source: The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

题意：

用 n 块面积为 n 的图块拼满一个 n*n 的矩阵图，并且 n 块图只有四种颜色，相邻的图块不能有相同的颜色，任意的两块不能有相同的形状，包括旋转，对称；

注意当n为6 的时候会有相同的块，所以要特判一下；

代码如下：

#include <cstdio>
#include <cstring>

const int MAXN = 117;
char mm[MAXN][MAXN];
char col[2];
int n;

void solve()
{
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
            mm[i][j] = 'B';
        }
    }
    for(int i = 0; i < n; i++)
        mm[0][i] = 'Y';
    for(int i = 0; i < (n-1)/2; i++)
    {
        char c = col[i%2];
        for(int j = i+1; j < n; j++)
        {
            mm[j][i] = c;
        }
        for(int j = 1; j <= i+1; j++)
        {
            mm[j][i+1] = c;
        }
    }
    for(int i = (n-1)/2; i < n-1; i++)
    {
        char c = col[i%2];
        for(int j = i+2; j < n; j++)
        {
            mm[j][i] = c;
        }
        mm[i+2][i+1] = c;
        for(int j = 2; j <= i+2; j++)
        {
            mm[j][i+2] = c;
        }
    }
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
            printf("%c",mm[i][j]);
        }
        printf("\n");
    }
}

int main()
{
    col[0] = 'G';
    col[1] = 'R';
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        if(n == 1)
            printf("Y\n");
        else if(n <= 4)
            printf("No solution!\n");
        else if(n == 6)//特判
        {
            printf("YYYYYY\n");
            printf("GGRGRR\n");
            printf("GRRGRB\n");
            printf("GRGGRB\n");
            printf("GRGRRB\n");
            printf("GRGBBB\n");
        }
        else
        {
            solve();
        }
    }
    return 0;
}