poj2676(数独 搜索)

题目链接

题意

解出N个9x9的数独,(输出一个解即可)

解题思路

用一个二维布尔数组来标记,点(x,y)的行,列,块的数字是否已经出现过,然后直接暴力搜索即可

AC代码

#include<vector>
#include<algorithm>
#include<cstdio>
#include<iostream>
#include<set>
#include<cstring>
#include<functional>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<int,pii> PII;
bool vis[100][10];
int G[10][10];
int N = 10;
int M = 20;

inline int getid(int& x,int& y)     //得到(x,y)所属的块
{
    if(x<3&&y<3) return M+1;
    else if(x<6&&y<3) return M+2;
    else if(x<9&&y<3)return M+3;
    else if(x<3&&y<6)return M+4;
    else if(x<6&&y<6) return M+5;
    else if(x<9&&y<6) return M+6;
    else if(x<3&&y<9)return M+7;
    else if(x<6&&y<9) return M+8;
    else return M+9;
}

bool ojbk = 0;

void dfs(int x,int y)
{
    if(ojbk)return;
    if(y>8){
        y %= 9;
        x++;
    }
    if(x==9&&y==0){
        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++){
                cout << G[i][j];
            }
        }
        ojbk = 1;
    }
    if(G[x][y]){
        dfs(x,y+1);
        return;
    }
    int r = y;
    int c = N+x;
    int id = getid(x,y);
    for(int k=1;k<=9;k++){
        if(vis[r][k]||vis[c][k]||vis[id][k]) continue;
        G[x][y] = k;
        vis[r][k] = 1;
        vis[c][k] = 1;
        vis[id][k] = 1;
        dfs(x,y+1);
        G[x][y] = 0;        //回朔
        vis[r][k] = 0;
        vis[c][k] = 0;
        vis[id][k] = 0;
    }
}


int main(int argc, char const *argv[])
{
    string s;
    while(cin >> s){
        if(s=="end")break;
        ojbk = 0;
        memset(vis,0,sizeof(vis));
        int k = 0;
        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++){
                if(s[k]=='.'){
                    G[i][j] = 0;
                    k++;
                    continue;
                }
                G[i][j] = s[k++]-'0';
                if(G[i][j]==0)continue;
                vis[j][G[i][j]] = 1;
                vis[N+i][G[i][j]] = 1;
                vis[getid(i,j)][G[i][j]] = 1;
            }
        }
        dfs(0,0);
        cout << endl;
    }
    return 0;
}
posted @ 2018-10-09 22:58  django_lf  阅读(252)  评论(0编辑  收藏  举报