ZOJ Problem Set - 1008 Gnome Tetravex (DFS+剪枝)

ZOJ Problem Set - 1008

 

Gnome Tetravex

 

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

 

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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

 

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Source: Asia 2001, Shanghai (Mainland China)

#include <iostream> #include <algorithm> #include <cstring>

using namespace std ; #define maxn 10000

struct node { int top , right, botom, left ; } ; int visit[maxn] ; node num[maxn] ; int Map[maxn] ; bool result_flag ; int n ; int a , b , c , d ; int total ; void DFS(int pos){ if(pos == n*n){ result_flag = true ; return; } if(result_flag){ return; } for(int i=0 ; i<total ; i++){ if(visit[i]){ if(pos == 0){ visit[i] -- ; Map[pos] = i ; DFS(pos+1) ; visit[i] ++ ; }else if(pos<n){ if(num[Map[pos-1]].right == num[i].left){ visit[i] -- ; Map[pos] = i ; DFS(pos+1) ; visit[i] ++ ; } }else if(pos%n == 0 ){ if(num[Map[pos-n]].botom == num[i].top){ visit[i] -- ; Map[pos] = i ; DFS(pos+1) ; visit[i] ++ ; } }else{ if(num[Map[pos-1]].right == num[i].left && num[Map[pos-n]].botom == num[i].top){ visit[i] -- ; Map[pos] = i ; DFS(pos+1) ; visit[i] ++ ; } } if(result_flag){ return; } } } return; } int main(){ int Case = 1 ; while(cin >> n && n ){ total = 0 ; memset(visit , 0 , sizeof(visit)) ; for(int i=0 ; i<n*n ; i++){ cin >> a >> b >> c >> d ; bool flag = false ; // 不加上会超时 // 因此必定要加上此处的剪枝
            for(int k=0 ; k<total ; k++){ if(num[k].top == a && num[k].right == b && num[k].botom == c && num[k].left == d){ visit[k] ++ ; flag = true ; break ; } } if(!flag){ num[total].top = a ; num[total].right = b ; num[total].botom = c ; num[total].left = d ; visit[total] ++ ; total ++ ; } } result_flag = false ; memset(Map , 0 , sizeof(Map)) ; DFS(0) ; if(Case != 1 ){ cout << endl ; } if(result_flag == true){ cout << "Game " << Case ++ << ": " << "Possible"<< endl ; }else{ cout << "Game " << Case ++ << ": " << "Impossible"<<endl ; } } return 0 ; }

/* 题意：给出一个矩形由n*n个小矩形，每一个小矩形由四个三角形组成，分别在上下左右，每一个三角形有一个数 字，经过调换这些矩形的位置，找出一种状况能使得任意两个相邻的小矩形之间有公共边的两个三角形的值 同样，能找出这张状况则输出possible，不然输出impossible。 题解：DFS,搜索剪枝; n*n个位置分别对应填入n*n个矩形，从左往右，自上而下的顺序依次搜索，对应每一个位置都遍历一遍全部的 矩形，而且记录该矩形的值。 注意：测试数据包含有重复的矩形，所以在搜索的时候若是没有处理重复的矩形，则会出现一种状况：遍历 的两种摆法中两个相同的矩形分别在两个位置，会由于没有处理重复而经过颠倒两个矩形的出现顺序来从新 搜索一遍，其实是同样的，这里的剪枝很重要，不然会超时，估计数据有不少重复的矩形。 */ #include <cstdio> #include <cstring> #include <algorithm>

using namespace std; struct square { int top,right,bottom,left; }s[50];//记录矩形

int n,q;//n记录大矩形的边，q记录矩形的种类数
int vis[50];//记录第i种矩形由多少个
int m[50];//记录n*n个位置中对应的矩形的种类
bool flag;//记录是否能找到相邻状态

void dfs(int pos)//pos指当前在位置
{ if (pos == n*n)//搜索到n*n使表示全部的位置都已经填满
 { flag = true; return ; } if (flag) return ; for(int i=0; i<q; i++) {//遍历全部存在的小矩形填入当前的pos位置中
        if (vis[i]) { if (pos==0)//位于第一个位置，则只需直接填入矩形i
 { vis[i]--;//第i种矩形数目减一
                m[pos] = i;//记录当前pos位置填入的矩形种类为i
                dfs(pos+1);//搜索下一个位置
                vis[i]++; } else if (pos < n)//当前位置位于第一行
 { if (s[m[pos-1]].right == s[i].left)//判断与左边是否能相邻(pos-1表示当前位置的左边)
 { vis[i]--; m[pos] = i; dfs(pos+1); vis[i]++; } } else if (pos % n == 0)//位于最左列，左边没有位置
 { if (s[m[pos-n]].bottom == s[i].top)//判断与上方的矩形是否能相邻(pos-n表示当前位置的上方)
 { vis[i]--; m[pos] = i; dfs(pos+1); vis[i]++; } } else {//其他的状况都判断是否能与上方和左边的矩形相邻
                if (s[m[pos-1]].right==s[i].left && s[m[pos-n]].bottom==s[i].top) { vis[i]--; m[pos] = i; dfs(pos+1); vis[i]++; } } if (flag) return ; } } } int main(void) { int cas=1,a,b,c,d; while (~scanf("%d",&n) && n) { memset(vis,0,sizeof(vis)); q = 0; for(int i=0; i<n*n; i++) { scanf("%d%d%d%d",&a,&b,&c,&d); int j; for(j=0; j<q; j++) { if (s[j].top==a && s[j].right==b && s[j].bottom==c && s[j].left==d) { vis[j]++; break; } } if (j == q) { s[q].top = a; s[q].right = b; s[q].bottom = c; s[q].left = d; vis[q]++; q++; } } flag = false; memset(m,0,sizeof(m)); dfs(0); if (cas != 1) printf("\n"); if (flag) printf("Game %d: Possible\n",cas); else printf("Game %d: Impossible\n",cas); cas++; } return 0; }

View Code

 

#include <iostream>#include <algorithm>#include <cstring>using namespace std ; #define maxn 10000struct node {    int top ,        right,        botom,        left ; } ; int visit[maxn] ; node num[maxn] ; int Map[maxn] ; bool result_flag ;int n ; int a , b , c , d ; int total ; void DFS(int pos){    if(pos == n*n){        result_flag = true ;         return;    }    if(result_flag){        return;    }    for(int i=0 ; i<total ; i++){        if(visit[i]){            if(pos == 0){                visit[i] -- ;                 Map[pos] = i ;                 DFS(pos+1) ;                 visit[i] ++ ;             }else if(pos<n){                if(num[Map[pos-1]].right == num[i].left){                    visit[i] -- ;                     Map[pos] = i ;                     DFS(pos+1) ;                     visit[i] ++ ;                 }            }else if(pos%n == 0 ){                if(num[Map[pos-n]].botom == num[i].top){                    visit[i] -- ;                     Map[pos] = i ;                     DFS(pos+1) ;                     visit[i] ++ ;                 }            }else{                if(num[Map[pos-1]].right == num[i].left && num[Map[pos-n]].botom == num[i].top){                    visit[i] -- ;                     Map[pos] = i ;                     DFS(pos+1) ;                     visit[i] ++ ;                 }            }            if(result_flag){                return;            }        }    }    return;}int main(){    int Case = 1 ;     while(cin >> n && n ){        total = 0 ;         memset(visit , 0 , sizeof(visit)) ;         for(int i=0 ; i<n*n ; i++){            cin >> a >> b >> c >> d ;            bool flag = false ;             //  不加上会超时              //  因此必定要加上此处的剪枝            for(int k=0 ; k<total ; k++){                if(num[k].top == a && num[k].right == b && num[k].botom == c && num[k].left == d){                    visit[k] ++ ;                     flag = true ;                     break ;                  }            }            if(!flag){                num[total].top = a ;                 num[total].right = b ;                 num[total].botom = c ;                 num[total].left = d ;                 visit[total] ++ ;                 total ++ ;             }        }        result_flag = false ;         memset(Map , 0 , sizeof(Map)) ;         DFS(0) ;        if(Case != 1 ){            cout << endl ;         }                if(result_flag == true){            cout << "Game " << Case ++ << ": " << "Possible"<< endl  ;         }else{            cout << "Game " << Case ++ << ": " << "Impossible"<<endl  ;        }    }    return 0 ; }