• UVA 211 The Domino Effect


    https://vjudge.net/problem/UVA-211

    题目

    给你28张“双六”多米诺骨牌,编号如下:

    Bone #      Pips        Bone #      Pips        Bone #      Pips        Bone #      Pips
       1        0 | 0          8        1 | 1         15        2 | 3         22        3 | 6
       2        0 | 1          9        1 | 2         16        2 | 4         23        4 | 4
       3        0 | 2         10        1 | 3         17        2 | 5         24        4 | 5
       4        0 | 3         11        1 | 4         18        2 | 6         25        4 | 6
       5        0 | 4         12        1 | 5         19        3 | 3         26        5 | 5
       6        0 | 5         13        1 | 6         20        3 | 4         27        5 | 6
       7        0 | 6         14        2 | 2         21        3 | 5         28        6 | 6
    

     在7*8的网格中每张牌各摆一张,如图所示,左边是各个格子的点数,右边是各个格子所属的骨牌编号。

      7 x 8 grid of pips            map of bone numbers
                                    
    6  6  2  6  5  2  4  1        28 28 14  7 17 17 11 11
    1  3  2  0  1  0  3  4        10 10 14  7  2  2 21 23
    1  3  2  4  6  6  5  4         8  4 16 25 25 13 21 23
    1  0  4  3  2  1  1  2         8  4 16 15 15 13  9  9
    5  1  3  6  0  4  5  5        12 12 22 22  5  5 26 26
    5  5  4  0  2  6  0  3        27 24 24  3  3 18  1 19
    6  0  5  3  4  2  0  3        27  6  6 20 20 18  1 19
    

     输入左图,你的任务是输出所有可能的右图

    题解

    整理数据= =将每张牌的每个数字与另外一边的数字对应起来,可以加快搜索速度

    只有28层,不用担心爆栈

    太激动了……又是花一大堆时间写的一道题!(4小时)

    (早睡早起.jpg)

    AC代码

    #include<bits/stdc++.h>
    using namespace std;
    #define REP(r,x,y) for(register int r=(x); r<(y); r++)
    #define REPE(r,x,y) for(register int r=(x); r<=(y); r++)
    #define MAXN 7
    #define MAXM 8
    #ifdef sahdsg
    #define DBG(...) printf(__VA_ARGS__)
    #else
    #define DBG(...)
    #endif
    int mp[MAXN][MAXM];
    int ans[MAXN][MAXM];
    bool vis[29];
    int ansc=0;
    int bones[][2] =
    {
    	{-1,-1},{0,0},{0,1},{0,2},{0,3},{0,4},{0,5},{0,6},{1,1},{1,2},{1,3},{1,4},{1,5},{1,6},{2,2},{2,3},{2,4},{2,5},{2,6},{3,3},{3,4},{3,5},{3,6},{4,4},{4,5},{4,6},{5,5},{5,6},{6,6}
    };
    int sq[][7]=
    {
    	{1*2,2*2+1,3*2+1,4*2+1,5*2+1,6*2+1,7*2+1},
    	{2*2,8*2,9*2+1,10*2+1,11*2+1,12*2+1,13*2+1},
    	{3*2,9*2,14*2,15*2+1,16*2+1,17*2+1,18*2+1},
    	{4*2,10*2,15*2,19*2,20*2+1,21*2+1,22*2+1},
    	{5*2,11*2,16*2,20*2,23*2,24*2+1,25*2+1},
    	{6*2,12*2,17*2,21*2,24*2,26*2,27*2+1},
    	{7*2,13*2,18*2,22*2,25*2,27*2,28*2}
    };
    inline bool getblock() {
    	ansc=0;
    	memset(ans,-1,sizeof ans);
    	REP(i,0,7) REP(j,0,8) {
    		if(scanf("%d", &mp[i][j])==-1) return false;
    	}
    	return true;
    }
    //int back;
    bool dfs(int d, int x, int y) {
    //	DBG("#%d	%d	%d
    ", d, x, y);
    	st:
    	if(y>=8) y=0, x++;
    	if(d>=28) {
    		ansc++;
    		REP(i,0,MAXN) {
    			REP(j,0,MAXM) {
    				if(j) putchar(' ');
    				printf("%d", ans[i][j]);
    			}
    			putchar('
    ');
    		}
    			putchar('
    ');
    		return true;
    	}
    	if(ans[x][y]!=-1) {
    		y++; goto st;
    	}
    	int k=mp[x][y];
    	bool f=false;
    	REP(i,0,8) {
    		int t=sq[k][i];
    		int p=t&1; t>>=1;
    		if(vis[t]) continue;
    //		assert(t>0);
    		if(x+1<7)
    		if(mp[x+1][y]==bones[t][p] && ans[x+1][y]==-1) {
    			f=true;vis[t]=true;
    			ans[x][y]=t, ans[x+1][y]=t;
    			dfs(d+1, x,y+1);
    			ans[x][y]=-1, ans[x+1][y]=-1;
    			vis[t]=false;
    		}
    		if(y+1<8)
    		if(mp[x][y+1]==bones[t][p] && ans[x][y+1]==-1) {
    			f=true;vis[t]=true;
    			ans[x][y]=t, ans[x][y+1]=t;
    			dfs(d+1, x,y+2);
    			ans[x][y]=-1, ans[x][y+1]=-1;
    			vis[t]=false;
    		}
    	}
    	return f;
    }
    
    int main() {
    	#ifdef sahdsg
    	freopen("in.txt", "r", stdin);
    	#endif
    	int kase=0;
    	while(getblock()) {
    		if(kase) puts("
    
    ");
    		kase++;
    		printf("Layout #%d:
    
    ", kase);
    		REP(i,0,7) {
    			REP(j,0,8) {
    				if(j) putchar(' ');
    				printf("%d", mp[i][j]);
    			}
    			putchar('
    ');
    		}
    		putchar('
    ');
    		printf("Maps resulting from layout #%d are:
    
    ", kase);
    		dfs(0,0,0);
    		printf("There are %d solution(s) for layout #%d.
    ", ansc, kase);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/sahdsg/p/10463763.html
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