DFS 题目，剪枝比较重要，这里使用的是把重复的方块只记录一次，用 num[] 保存它的数目。

# include 
     
      # include 
      
       # define N 25 + 2bool finished;int n, m, t[N][4], num[N], ans[N];void dfs(int cnt){    if (cnt == n*n) {finished = true; return ;}    int x = cnt/n + 1, y = cnt%n + 1;    int left = cnt, top = cnt+1-n;    for (int i = 0; i < m; ++i) if (num[i])    {        bool ok = true;        if (x!=1 && t[i][0] != t[ans[top]][2]) ok = false;        if(ok && y!=1 && t[i][3] != t[ans[left]][1]) ok = false;        if (ok)        {            --num[i];            ans[cnt+1] = i;            dfs(cnt+1);            if(finished) return ;            ++num[i];        }    }}void init(void){    bool rpt;    int top, rig, bot, lef;    m = 0;    memset(num, 0, sizeof(num));    for (int i = 0; i < n*n; ++i)    {        rpt = false;        scanf("%d%d%d%d", &top, &rig, &bot, &lef);        for (int j = 0; j < m; ++j)        {            if (t[j][0] == top && t[j][1] == rig &&                t[j][2] == bot && t[j][3] == lef)                {                    rpt = true;                    ++num[j];                    break;                }        }        if (rpt) continue;        t[m][0] = top, t[m][1] = rig, t[m][2] = bot, t[m][3] = lef;        ++num[m];        ++m;    }}void solve(void){    finished = false;    dfs(0);    if (finished) puts("Possible");    else puts("Impossible");}int main(){    int i = 0;    while (~scanf("%d", &n))    {        if (!n) break;        init();        if(i != 0) putchar('\n');        printf("Game %d: ", ++i);        solve();    }    return 0;}
      
     

/**/