poj--2676 Sudoku(dfs)

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Sudoku

题解

从左上角到右下角一行一行填数。
递归需要返回状态，即填数是否成功，如果成功就无需回溯，否则需要回溯。

#include <iostream>#include <cstdio>#include <string>#include <algorithm>#include <cstring>using namespace std;#define MEM(a, v) memset(a, v, sizeof(a))const int N = 9;int a[N+1][N+1];int rowFlag[N+1][N+1], colFlag[N+1][N+1], squFlag[N+1][N+1];int t;string s;bool Sudoku(int r, int c){    if(r > N) return true;    bool flag = false;    if(a[r][c]){        if(c == N)  flag = Sudoku(r + 1, 1);        else flag = Sudoku(r, c + 1);        return flag;    }    for(int i = 1; i <= N; ++i){        int k = 3*((r-1)/3)+(c-1)/3 + 1;        if(!rowFlag[r][i] && !colFlag[c][i] && !squFlag[k][i]){            rowFlag[r][i] = colFlag[c][i] = squFlag[k][i] = 1;            a[r][c] = i;            if(c == N) flag = Sudoku(r + 1, 1);            else flag = Sudoku(r, c + 1);            if(flag) return true;   // 填数成功不回溯            a[r][c] = 0;            rowFlag[r][i] = colFlag[c][i] = squFlag[k][i] = 0;        }    }    return false;}int main(){#ifdef EXMYfreopen("data.in", "r", stdin);#endif // EXMY    /// 712K    438MS    cin >> t;    while(t--){        MEM(rowFlag, 0);        MEM(colFlag, 0);        MEM(squFlag, 0);        for(int i = 1; i <= N; ++i){            cin >> s;            for(int j = 1; j <= s.length(); ++j){                int t = s[j-1] - '0';                a[i][j] = t;                if(a[i][j]){                    int k = 3*((i-1)/3)+(j-1)/3 + 1;                    rowFlag[i][t] = colFlag[j][t] = squFlag[k][t] = 1;                }            }        }        Sudoku(1, 1);        for(int i = 1; i <= N; ++i){            for(int j = 1; j <= N; ++j) cout << a[i][j];            cout << endl;        }    }    return 0;}

还可以更简洁：

#include <iostream>#include <cstdio>#include <string>#include <algorithm>#include <cstring>using namespace std;#define MEM(a, v) memset(a, v, sizeof(a))const int N = 9;int a[N+1][N+1];int rowFlag[N+1][N+1], colFlag[N+1][N+1];int squFlag[3][3][N+1];int t;string s;bool Sudoku(int k){    if(k >= N * N) return true;    int r = k / 9;    int c = k % 9;    if(a[r][c]) return Sudoku(k + 1);    for(int i = 1; i <= N; ++i){        if(!rowFlag[r][i] && !colFlag[c][i] && !squFlag[r/3][c/3][i]){            rowFlag[r][i] = colFlag[c][i] = squFlag[r/3][c/3][i] = 1;            a[r][c] = i;            if(Sudoku(k + 1)) return true;            rowFlag[r][i] = colFlag[c][i] = squFlag[r/3][c/3][i] = 0;            a[r][c] = 0;        }    }    return false;}int main(){#ifdef EXMYfreopen("data.in", "r", stdin);#endif // EXMY    cin >> t;    while(t--){        MEM(rowFlag, 0);        MEM(colFlag, 0);        MEM(squFlag, 0);        for(int i = 0; i < N; ++i){            cin >> s;            for(int j = 0; j < s.length(); ++j){                int t = s[j] - '0';                a[i][j] = t;                if(a[i][j]){                    rowFlag[i][t] = colFlag[j][t] = squFlag[i/3][j/3][t] = 1;                }            }        }        Sudoku(0);        for(int i = 0; i < N; ++i){            for(int j = 0; j < N; ++j) cout << a[i][j];            cout << endl;        }    }}

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