poj 3074 DLX模板

// 题目链接 http://poj.org/problem?id=3074

 // 精确覆盖问题求解转载：https://www.cnblogs.com/grenet/p/3145800.html

// 把数独问题转换为精确覆盖问题转载：https://www.cnblogs.com/grenet/p/3163550.html

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;

//const int INF = 1 << 30;
const int maxn = 729;
const int maxnode = 2916;
const int maxnr = 729;

//行编号从1开始,列编号为1~n,结点0是表头结点;结点1~n是各列顶部的虚拟结点
struct DLX {
    int n, sz;//列数, 结点总数
    int S[maxn];//各列节点数

    int row[maxnode], col[maxnode];//各结点行列编号
    int L[maxnode], R[maxnode], U[maxnode], D[maxnode];//十字链表

    int ansd, ans[maxnr];//解

    void init(int n) {//n是列数
        this->n = n;

        //虚拟结点
        for(int i = 0; i <= n; ++i) {
            U[i] = i;
            D[i] = i;
            L[i] = i-1;
            R[i] = i+1;
        }
        R[n] = 0; L[0] = n;

        sz = n + 1;
        memset(S, 0, sizeof(S));
    }

    void addRow(int r, vector<int> columns) {
        int first = sz;
        for(int i = 0; i != columns.size(); ++i) {
            int c = columns[i];
            L[sz] = sz-1; R[sz] = sz+1; D[sz] = c; U[sz] = U[c];
            D[U[c]] = sz; U[c] = sz;
            row[sz] = r; col[sz] = c;
            S[c]++; sz++;
        }
        R[sz-1] = first; L[first] = sz-1;
    }
    // 顺着链表A, 遍历除s外的其它元素
    #define FOR(i, A, s) for(int i = A[s]; i != s; i = A[i])

    void remove(int c) {
        L[R[c]] = L[c];
        R[L[c]] = R[c];
        FOR(i, D, c)
            FOR(j, R, i) {
                U[D[j]] = U[j];
                D[U[j]] = D[j];
                --S[col[j]];
            }
    }

    void restore(int c) {
        FOR(i, U, c) 
            FOR(j, L, i) {
                ++S[col[j]];
                U[D[j]] = j;
                D[U[j]] = j;
            }
        L[R[c]] = c;
        R[L[c]] = c;
    }

    // d为递归深度
    bool dfs(int d) {
        if(R[0] == 0) {  // 找到解
            ansd = d;    // 记录解的长度
            return true;
        }
        
        // 找s最小的列c
        int c = R[0];              // 第一个未删除的列
        FOR(i, R, 0) if(S[i] < S[c]) c = i;

        remove(c);                 // 删除c列
        FOR(i, D, c) {             // 用结点i所在的行覆盖第c列
            ans[d] = row[i];
            FOR(j, R, i) remove(col[j]);// 删除结点i所在行能覆盖的所有其它列
            if(dfs(d+1)) return true;
            FOR(j, L, i) restore(col[j]);// 恢复结点i所在行能覆盖的所有其它列
        }
        restore(c);                // 恢复第c列

        return false;
    }

    bool solve(vector<int>& v) {
        v.clear();
        if(!dfs(0)) return false;
        for(int i = 0; i != ansd; ++i) v.push_back(ans[i]);
        return true;
    }
};

const int SLOT = 0;
const int ROW = 1;
const int COL = 2;
const int SUB = 3;

inline int encode(int a, int b, int c) {
    return a*81 + b*9 + c + 1;
}

inline void decode(int code, int &a, int &b, int &c) {
    code--;
    c = code%9; code /= 9;
    b = code%9; code /= 9;
    a = code;
}

DLX solver;

int main() {
    char str[81];
    char puzzle[9][9];
    while(scanf("%s", str) == 1 && strcmp(str, "end")) {
        solver.init(324);
        for(int i = 0; i != 9; ++i) {
            for(int j = 0; j != 9; ++j) {
                puzzle[i][j] = str[i*9 + j];
            }
        }
        for(int r = 0; r != 9; ++r) {
            for(int c = 0; c != 9; ++c) {
                for(int v = 0; v != 9; ++v) {
                    if(puzzle[r][c] == '.' || puzzle[r][c]-'0' == v+1) {
                        vector<int> columns;
                        columns.push_back(encode(SLOT, r, c));
                        columns.push_back(encode(ROW, r, v));
                        columns.push_back(encode(COL, c, v));
                        columns.push_back(encode(SUB, (r/3)*3 + c/3, v));
                        solver.addRow(encode(r, c, v), columns);
                    }
                }
            }
        }
        vector<int> ans;
        solver.solve(ans);

        for(int i = 0; i != ans.size(); ++i) {
            int r, c, v;
            decode(ans[i], r, c, v);
            puzzle[r][c] = '1'+v;
        }
        for(int i = 0; i != 9; ++i) {
            for(int j = 0; j != 9; ++j) {
                printf("%c", puzzle[i][j]);
            }
        }
        printf("
");
    }
    return 0;
}