ZOJ 1008 Gnome Tetravex(DFS)

Gnome Tetravex

---

Time Limit: 10 Seconds      Memory Limit: 32768 KB

---

Hart is engaged in playing an interesting game, Gnome Tetravex, these days. In the game, at the beginning, the player is given n*n squares. Each square is divided into four triangles marked four numbers (range from 0 to 9). In a square, the triangles are the left triangle, the top triangle, the right triangle and the bottom triangle. For example, Fig. 1 shows the initial state of 2*2 squares.

Fig. 1 The initial state with 2*2 squares

The player is required to move the squares to the termination state. In the
termination state, any two adjoining squares should make the adjacent triangle
marked with the same number. Fig. 2 shows one of the termination states of the
above example.

Fig. 2 One termination state of the above example

It seems the game is not so hard. But indeed, Hart is not accomplished in the
game. He can finish the easiest game successfully. When facing with a more complex
game, he can find no way out.

One day, when Hart was playing a very complex game, he cried out, "The
computer is making a goose of me. It's impossible to solve it." To such
a poor player, the best way to help him is to tell him whether the game could
be solved. If he is told the game is unsolvable, he needn't waste so much time
on it.

Input

The input file consists of several game cases. The first line of each game case
contains one integer n, 0 <= n <= 5, indicating the size of the game.

The following n*n lines describe the marking number of these triangles. Each
line consists of four integers, which in order represent the top triangle, the
right triangle, the bottom triangle and the left triangle of one square.

After the last game case, the integer 0 indicates the termination of the input
data set.

Output

You should make the decision whether the game case could be solved. For each
game case, print the game number, a colon, and a white space, then display your
judgment. If the game is solvable, print the string "Possible". Otherwise,
please print "Impossible" to indicate that there's no way to solve
the problem.

Print a blank line between each game case.

Note: Any unwanted blank lines or white spaces are unacceptable.

Sample Input

2
5 9 1 4
4 4 5 6
6 8 5 4
0 4 4 3
2
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
0

Output for the Sample Input

Game 1: Possible

Game 2: Impossible

题意：在一个N*N矩形区域中，N*N个小矩形重新拼成一个符合规则（左右相连，上下相连值相等）的矩形。

思路：判断出有多种小矩形，接着一个一拼。

收获：异或：两者不相等时为1.

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
using namespace std;

const int INF=0x3f3f3f3f;
const double eps=1e-;
const double PI=acos(-1.0);
#define maxn 30

struct Node
{
    int u, r, d, l;
};
Node node[maxn];
int num[maxn];
int map1[][];
int n, cnt;
bool fk;
void dfs(int p)
{
    if(fk) return;
    if(p == n*n)
    {
        fk = ;
        return;
    }
    for(int i = ; i < cnt; i++)
    {
        if(num[i] == ) continue;
        int x = p/n;
        int y = p%n;
        if(y> && node[map1[x][y-]].r ^ node[i].l) continue;
        if(x> && node[map1[x-][y]].d ^ node[i].u) continue;
        map1[x][y] = i;
        num[i]--;
        dfs(p+);
        if(fk) return ;
        num[i]++;
    }
}
int main()
{
    int cas = ;
    int u, r, l, d;
    int flag = ;
    while(~scanf("%d",&n) && n)
    {
        cnt = ;
        memset(num, , sizeof num);
        for(int i = ; i < n*n; i++)
        {
            scanf("%d%d%d%d",&u, &r, &d, &l);
            int f = ;
            for(int j = ; j < cnt; j++)
            {
                if(node[j].d == d && node[j].l == l && node[j].r == r && node[j].u == u)
                    {
                        num[j]++;
                        f = ;
                        break;
                    }
            }
            if(!f)
            {
            node[cnt].d = d; node[cnt].l = l; node[cnt].r = r; node[cnt].u = u;
            num[cnt] = ;
            cnt++;
            }
        }
        fk = ;
        dfs();
        if(cas > )
            puts("");
        printf("Game %d: ",cas++);
        if(fk)
            printf("Possible\n");
        else
            printf("Impossible\n");
    }
    return ;
}