UVA - 211 The Domino Effect（多米诺效应）（dfs回溯）

题意：根据多米诺骨牌的编号的7*8矩阵，每个点可以和相邻的点组成的骨牌对应一个编号，问能形成多少种由编号组成的图。

分析：dfs，组成的图必须有1~28所有编号。

#pragma comment(linker, "/STACK:102400000, 102400000")

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define Min(a, b) ((a < b) ? a : b)

#define Max(a, b) ((a < b) ? b : a)

typedef long long ll;

typedef unsigned long long llu;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {, };

const int dc[] = {, };

const int MOD = 1e9 + ;

const double pi = acos(-1.0);

const double eps = 1e-;

const int MAXN =  + ;

const int MAXT =  + ;

using namespace std;

int a[][];

int ans[][];

int id[][];//多米诺骨牌的编号

int vis[];

int mark[][];//是否访问过该点

int cnt;

void init(){

    int k = ;

    for(int i = ; i < ; ++i){

        for(int j = i; j < ; ++j){

            id[i][j] = id[j][i] = ++k;

        }

    }

}

bool judge1(int x, int y){

    return x >=  && x <=  && y >=  && y <= ;

}

bool judge2(int x){

    for(int i = ; i < ; ++i){

        if(!mark[x][i]) return true;

    }

    return false;

}

void dfs(int x, int y, int cur){

    if(cur == ){

        ++cnt;

        for(int i = ; i < ; ++i){

            for(int j = ; j < ; ++j){

                printf("%4d", ans[i][j]);

            }

            printf("\n");

        }

        printf("\n");

        return;

    }

    if(x == ) return;

    if(y == ){

        if(judge2(x)) return;//如果这一行有未访问的

        dfs(x + , , cur);

        return;

    }

    if(mark[x][y]){

        dfs(x, y + , cur);

        return;

    }

    for(int i = ; i < ; ++i){//向右或向下

        int tx = x + dr[i];

        int ty = y + dc[i];

        if(judge1(tx, ty) && !mark[tx][ty] && !vis[id[a[x][y]][a[tx][ty]]]){//下标合法、未被访问过，编号未用过

            ans[x][y] = ans[tx][ty] = id[a[x][y]][a[tx][ty]];

            mark[x][y] = mark[tx][ty] = vis[id[a[x][y]][a[tx][ty]]] = ;

            dfs(x, y + , cur + );

            mark[x][y] = mark[tx][ty] = vis[id[a[x][y]][a[tx][ty]]] = ;

        }

    }

}

int main(){

    init();

    int x;

    int kase = ;

    while(scanf("%d", &x) == ){

        if(kase) printf("\n\n\n");

        cnt = ;

        memset(vis, , sizeof vis);

        memset(ans, , sizeof ans);

        memset(mark, , sizeof mark);

        a[][] = x;

        for(int i = ; i < ; ++i){

            for(int j = ; j < ; ++j){

                if(!i && !j) continue;

                scanf("%d", &a[i][j]);

            }

        }

        printf("Layout #%d:\n\n", ++kase);

        for(int i = ; i < ; ++i){

            for(int j = ; j < ; ++j){

                printf("%4d", a[i][j]);

            }

            printf("\n");

        }

        printf("\nMaps resulting from layout #%d are:\n\n", kase);

        dfs(, , );

        printf("There are %d solution(s) for layout #%d.\n", cnt, kase);

    }

    return ;

}