UVa 211 - The Domino Effect (DFS)

题意

多米诺骨牌有28种型号，每种型号都由两个pip组成。

比如bone 1 是由 0和0组成的，bone 18是由2和6组成的。

现在给出一个pip组成的图，要求输出所有能组成的bone型号排列。

我会说我看懂题目花了一个小时吗TAT

对于每种型号的骨牌，要么横着放，要么竖着放。一直往右DFS + 回溯即可。

代码

#include <stack>
#include <cstdio>
#include <list>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 4e5 + 10;
const int MOD = 1000007;
const int dir[][2] = { {1, 0}, {0, 1} };
int cases = 0;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
const int row = 7, col = 8;
int mp[row][col], cvt[row][col], vis[30], pip[row][col];
int cnt;
bool Input()
{
    for (int i = 0; i < row; i++)
        for (int j = 0; j < col; j++) if (scanf("%d", &pip[i][j]) == -1) return false;
    return true;
}
void Init()
{
    int num = 1;
    cnt = 0;
    for (int i = 0; i < row; i++)
        for (int j = i; j < row; j++) cvt[i][j] = cvt[j][i] = num++;
}
void Output(int p[7][8])
{
    for (int i = 0; i < row; i++)
    {
        for (int j = 0; j < col; j++) printf("%4d", p[i][j]);
        puts("");
    }
    puts("");
}
void DFS(int x, int y)
{
    if (y == col)
    {
        x++, y = 0;
        if (x == row)
        {
            cnt++;
            Output(mp);
            return;
        }
    }
    if (mp[x][y]) DFS(x, y + 1);
    else
    {
        for (int i = 0; i < 2; i++)
        {
            int xx = x + dir[i][0], yy = y + dir[i][1];
            if (xx < row && yy < col && !mp[xx][yy])
            {
                int bone = cvt[pip[x][y]][pip[xx][yy]];
                if (vis[bone]) continue;
                mp[x][y] = mp[xx][yy] = bone;
                vis[bone] = 1;
                DFS(x, y + 1);
                vis[bone] = 0;
                mp[x][y] = mp[xx][yy] = 0;
            }
        }
    }
}
int main()
{
    //ROP;
    Init();
    while (1)
    {
        cnt = 0;
        if (!Input()) break;
        if (cases) printf("\n\n\n");
        printf("Layout #%d:\n\n", ++cases);
        Output(pip);
        printf("Maps resulting from layout #%d are:\n\n", cases);
        DFS(0, 0);
        printf("There are %d solution(s) for layout #%d.\n", cnt, cases);
    }
    return 0;
}