题目链接：zoj 3810 A Volcanic Island

题目大意：给定n，要求用n块面积为n的拼图铺满n?n的矩阵，任意两块拼图形状不能相同（包括旋转和镜像），并且n块拼图只能有4中颜色，相邻两块拼图颜色不能相同。

解题思路：构造，n = 2，3，4时是不存在的。然后对于n >= 5的直接构造，具体看代码。注意这种构造方式构造6的时候会出现相同的拼图，所以特判。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 105;
const char s[10][10] = {"BBBBBB", "GGRGRR", "GRRGRB", "GRGGRB", "GRGRRB", "GRGBBB"};
const char c[5] = "BGRY";

int g[maxn][maxn];

void solve (int n) {
    memset(g, 0, sizeof(g));

    for (int i = 0; i < n; i++)
        g[0][i] = 3;

    int k = (n - 1) / 2, col = 1;
    for (int i = 0; i < k; i++) {
        for (int j = 1; j <= i+1; j++)
            g[j][i+1] = col;
        for (int j = i+1; j < n; j++)
            g[j][i] = col;
        col = 3 - col;
    }

    for (int i = k; i < n; i++) {
        for (int j = 2; j <= i+2; j++)
            g[j][i+2] = col;
        g[i+2][i+1] = col;
        for (int j = i+2; j < n; j++)
            g[j][i] = col;
        col = 3 - col;
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            printf("%c", c[g[i][j]]);
        printf("\n");
    }
}

int main () {
    int cas, n;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d", &n);
        if (n == 1)
            printf("B\n");
        else if (n == 6) {
            for (int i = 0; i < 6; i++)
                printf("%s\n", s[i]);
        } else if (n >= 5) {
            solve(n);
        } else
            printf("No solution!\n");
    }
    return 0;
}

zoj 3810 A Volcanic Island(构造)