搜索_常规DFS_位运算_HDOJ5547_Sudoku

Problem Description

Yi Sima was one of the best counselors of Cao Cao. He likes to play a funny game himself. It looks like the modern Sudoku, but smaller.

Actually, Yi Sima was playing it different. First of all, he tried to generate a 4×44×4 board with every row contains 1 to 4, every column contains 1 to 4. Also he made sure that if we cut the board into four 2×22×2 pieces, every piece contains 1 to 4.

Then, he removed several numbers from the board and gave it to another guy to recover it. As other counselors are not as smart as Yi Sima, Yi Sima always made sure that the board only has one way to recover.

Actually, you are seeing this because you've passed through to the Three-Kingdom Age. You can recover the board to make Yi Sima happy and be promoted. Go and do it!!!

Input

The first line of the input gives the number of test cases, T(1≤T≤100) . T test cases follow. Each test case starts with an empty line followed by 4 lines. Each line consist of 4 characters. Each character represents the number in the corresponding cell (one of '1', '2', '3', '4'). '*' represents that number was removed by Yi Sima.

It's guaranteed that there will be exactly one way to recover the board.

Output

For each test case, output one line containing Case #x:, where x is the test case number (starting from 1). Then output 4 lines with 4 characters each. indicate the recovered board.

思路分析:

使用DFS, 每次选择可填方案数最少的格子填数即可, 下面给出基于此策略的AC代码:

//HDOJ5547_Sudoku
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX = 100, INF = 0x3f3f3f3f;
const int ma[5][5] = {{0, 0, 0, 0, 0}
                    , {0, 1, 1, 2, 2}
                    , {0, 1, 1, 2, 2}
                    , {0, 3, 3, 4, 4}
                    , {0, 3, 3, 4, 4}};
char G[MAX][MAX];
int R[5], C[5], M[5];//R, C, M分别为行列宫格可选元素 
int lowbit(int num){
    int res = 0; while(num) num -= num & -num, ++res; return res;
}
//返回可选方案数最少的位置, (0, 0)表示所有数字均已填好, (-1, -1)表示不存在合法位置 
pair<int, int> getNex(){
    int resx = 0, resy = 0, cnt = INF;
    for(int i = 1; i <= 4; ++i)
        for(int j = 1; j <= 4; ++j)
            if(G[i][j] == '*'){
                int y = R[i] & C[j] & M[ma[i][j]], t = lowbit(y);
                if(!t) return make_pair(-1, -1);
                if(t < cnt) cnt = t, resx = i, resy = j;        
            }
    return make_pair(resx, resy);    
} 
bool dfs(int x, int y){
    int choosn = R[x] & C[y] & M[ma[x][y]];
    for(int i = 1; i <= 4; ++i)
        if(choosn >> i - 1 & 1){
            G[x][y] = '0' + i, R[x] ^= 1 << i - 1, C[y] ^= 1 << i - 1
			, M[ma[x][y]] ^= 1 << i - 1;
            pair<int, int> nex = getNex(); 
            if(nex.first == -1) return false;
            if(!nex.first) return true;
            if(dfs(nex.first, nex.second)) return true;
            G[x][y] = '*', R[x] ^= 1 << i - 1, C[y] ^= 1 << i - 1
			, M[ma[x][y]] ^= 1 << i - 1;
        }
    return false;
}
int main(){
    int T, sn = 0; scanf("%d", &T);
    while(++sn, T--){
        memset(R, 0x3f, sizeof(R)), memset(C, 0x3f, sizeof(C)), memset(M, 0x3f, sizeof(M));
        for(int i = 1; i <= 4; ++i) scanf("%s", G[i] + 1);
        for(int i = 1; i <= 4; ++i)
            for(int j = 1; j <= 4; ++j)
                if(G[i][j] != '*'){
                    int y = G[i][j] - '0' - 1;
                    R[i] ^= 1 << y, C[j] ^= 1 << y, M[ma[i][j]] ^= 1 << y; 
                }
        pair<int, int> pos = getNex();
        dfs(pos.first, pos.second);
        cout << "Case #" << sn << ":" << endl;
        for(int i = 1; i <= 4; ++i) cout << &G[i][1] << endl; 
    }
    return 0;
}