UVA 211 The Domino Effect 多米诺效应 （回溯）

骨牌无非两种放法，横着或竖着放，每次检查最r，c最小的没访问过的点即可。如果不能放就回溯。

最外面加一层认为已经访问过的位置，方便判断。

#include<bits/stdc++.h>



const int MAXD = 56;

const int MAXB = 29;

const int MAXP = 7;



bool used[MAXB];// used Bone

int pip[MAXP][MAXP];// pip 2 Bone



int tab[7][8];

int ans[8][9];



int cnt;

void dfs(int r,int c)

{

    while(ans[r][c]) {

        if(c == 8) { r++; c = -1; }

        c++;

    }

    if(r == 7) {

        for(int i = 0; i < 7; i++){

        for(int j = 0; j < 8; j++){

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

        }

        putchar('\n');

    }

    putchar('\n');

        cnt++;

        return ;

    }



    int &p0 = tab[r][c];

    if(!ans[r][c+1]){ // 横着放

        int &p1 = tab[r][c+1];

        int &bone = pip[p0][p1];

        if(!used[bone]){

            used[bone] = 1;

            ans[r][c+1] = ans[r][c] = bone;

            dfs(r,c+1);

            ans[r][c+1] = ans[r][c] = 0;

            used[bone] = 0;

        }

    }



    if(!ans[r+1][c]) { //竖着放

         int &p2 = tab[r+1][c];

         int &bone = pip[p0][p2];

         if(!used[bone]){

            used[bone] = 1;

            ans[r+1][c] = ans[r][c] = bone;

            dfs(r,c+1);

            ans[r+1][c] = ans[r][c] = 0;

            used[bone] = 0;

         }

    }



}



int main()

{

  //  freopen("in.txt","r",stdin);

   // freopen("my.txt","w",stdout);

    for(int h = 0,c = 0; h < 7; h++)

    for(int i = h; i < 7; i++){

        pip[i][h] = pip[h][i] = ++c;

    }

    int *layout = *tab;

    int cas = 0;

    while(~scanf("%d",layout)){

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

        memset(ans,0,sizeof(ans));

        memset(used,0,sizeof(used));

        cnt = 0;

        for(int i = 0; i < 7; i++) ans[i][8] = 233;

        for(int i = 0; i < 8; i++) ans[7][i] = 233;



        for(int i = 1; i < MAXD; i++)

            scanf("%d",layout+i);

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

        for(int i = 0; i < 7; i ++){

            for(int j = 0; j < 8; j++)

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

            putchar('\n');

        }

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

        dfs(0,0);

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

    }



    return 0;

}