211:The Domino Effect

The Domino Effect

回溯加剪枝。只需DFS右、下兩個方向即可，可以一行一行來。弄完一行之後看下該行所有數字是否都配對了，否則剪枝。還是得細心，兩個 cnt 認錯了找了大半天 bug 。。。

#include<bits/stdc++.h>
using namespace std;
const int n = 7;
int cnt;
int dx[] = {1, 0}, dy[] = {0, 1};
int dict[n][n], G[n][n+1], M[n][n+1], vis[n][n+1], use[30];

bool judge(int x) {
    for (int j = 0; j <= n; j++)
        if (vis[x][j] == 0) return true;
    return false;
}

void DFS(int x, int y, int cnt2){
    if(cnt2 == 28){
        for(int i = 0; i < n; i++){
            for(int j = 0; j <= n; j++){
                printf("%4d", M[i][j]);
            }
            putchar('\n');
        }
        putchar('\n');
        cnt++;
        return;
    }
    if(x == n) return;
    if(y == n+1){
        if (judge(x)) return;
        DFS(x + 1, 0, cnt2);
        return;
    }
    if(vis[x][y]){
        DFS(x, y + 1, cnt2);
        return;
    }
    for(int i = 0; i < 2; i++){
        int nx = x + dx[i], ny = y + dy[i];
        int k = dict[G[x][y]][G[nx][ny]];
        if(nx == n || ny == n+1 || vis[nx][ny] || use[k]) continue;
        M[x][y] = M[nx][ny] = k;
        vis[x][y] = vis[nx][ny] = use[k] = 1;
        DFS(x, y + 1, cnt2 + 1);
        vis[x][y] = vis[nx][ny] = use[k] = 0;
    }
}

int main()
{
    // freopen("data.in","r",stdin);
    // freopen("data.out","w",stdout);
    int c = 1;
    for(int i = 0; i < n; i++)
        for(int j = i; j < n; j++)
            dict[j][i] = dict[i][j] = c++;

    int T = 0;
    while(scanf("%d", &G[0][0]) != EOF){
        for(int i = 0; i < n; i++){
            for(int j = 0; j <= n; j++){
                if(!i && !j) continue;
                scanf("%d", &G[i][j]);
            }
        }
        if(T++) printf("\n\n\n");
        printf("Layout #%d:\n\n", T);
        for(int i = 0; i < n; i++){
            for(int j = 0; j <= n; j++){
                printf("%4d", G[i][j]);
            }
            putchar('\n');
        }
        printf("\nMaps resulting from layout #%d are:\n\n", T);
        cnt = 0;
        DFS(0, 0, 0);
        printf("There are %d solution(s) for layout #%d.\n", cnt, T);
    }

    return 0;
}