【习题 7-3 UVA - 211】The Domino Effect

【链接】 我是链接,点我呀:)
【题意】

在这里输入题意

【题解】

每次搜素要往下还是要往右摆。 然后维护一下让每个下标只出现一次就可以了。 =>作为剪枝条件

【代码】

/*
  	1.Shoud it use long long ?
  	2.Have you ever test several sample(at least therr) yourself?
  	3.Can you promise that the solution is right? At least,the main ideal
  	4.use the puts("") or putchar() or printf and such things?
  	5.init the used array or any value?
  	6.use error MAX_VALUE?
  	7.use scanf instead of cin/cout?
  	8.whatch out the detail input require
*/
/*
    一定在这里写完思路再敲代码!!!
    爆搜。
    把每一行、每一列都排满。
    然后枚举当前格子是往下排还是往右排
*/
#include <bits/stdc++.h>
using namespace std;

const int N = 10;

int a[N][N];
int bo[N][N],now[N][N];
int cnt[100],ans;

void Init(){
    int cnt = 1;
    for (int i = 0;i <= 6;i++){
        for (int j = i;j <= 6;j++){
            bo[i][j] = cnt++;
        }
    }
}

bool out(){
    for (int i = 1;i <=7;i++){
        for (int j = 1;j <= 8;j++)
            cout <<setw(4)<<now[i][j];
        cout << endl;
    }
    cout << endl << endl;
    return true;
}

void dfs(int x,int y){
    if (y==9){
        if (x==7){
            ans++;
            out();
            return;
        }
        dfs(x+1,1);
        return;
    }
    if (now[x][y]!=0){
        dfs(x,y+1);
        return;
    }
    //向下
    if (x+1<=7){
        int tx = min(a[x+1][y],a[x][y]),ty = max(a[x+1][y],a[x][y]);
        now[x+1][y] = now[x][y] = bo[tx][ty];
        cnt[bo[tx][ty]]++;
        if (cnt[bo[tx][ty]]==1)dfs(x,y+1);
        cnt[bo[tx][ty]]--;
        now[x+1][y] = now[x][y] = 0;
    }
    //向右
    if (y+1<=8 && now[x][y+1]==0){
        int tx = min(a[x][y+1],a[x][y]),ty = max(a[x][y+1],a[x][y]);
        now[x][y+1] = now[x][y] = bo[tx][ty];
        cnt[bo[tx][ty]]++;
        if (cnt[bo[tx][ty]]==1)dfs(x,y+1);
        cnt[bo[tx][ty]]--;
        now[x][y+1] = now[x][y] = 0;
    }
}

int main(){
	#ifdef LOCAL_DEFINE
	    freopen("rush_in.txt", "r", stdin);
	#endif
	ios::sync_with_stdio(0),cin.tie(0);
	Init();
	int Kase = 0;
    while (cin >> a[1][1]){
        if (Kase>0){
            cout << endl<<endl<<endl<<endl<<endl;
        }
        ans = 0;
        for (int j = 2;j <= 8;j++) cin >> a[1][j];
        for (int i = 2;i <= 7;i++)
            for (int j = 1;j <= 8;j++)
                cin >> a[i][j];

        cout <<"Layout #"<<++Kase<<":"<<endl<<endl<<endl;
        for (int i = 1;i <= 7;i++){
            for (int j = 1;j <= 8;j++){
                cout <<setw(3)<<a[i][j];
            }
            cout << endl;
        }
        cout << endl;
        cout <<"Maps resulting from layout #"<<Kase<<" are:"<<endl<<endl<<endl;
        dfs(1,1);
        cout <<"There are "<<ans<<" solution(s) for layout #"<<Kase<<"."<<endl;
    }
	return 0;
}