PKU 3074

    DLX數獨建圖：一共9*9*9 = 729行，9*9*4 = 324列。第i行表示數獨的i/81行i/9%9列放置數字i%9。324列分成4個部分，每個部分81列，分別限制每個格子只能放一個數字、每行只能放一種數字、每列只能放一種數字、每個3*3的格子只能放一種數字。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <iostream>
#include <sstream>

using namespace std;

const int maxn = 3241;
const int inf  = 1000000000;

char mtx[729][324];
char sudoku[82];
int N, M, head, idx;                        // 分別是行數, 列數, 列頭結點鏈表的頭結點, 結點計數器
int L[maxn], R[maxn], U[maxn], D[maxn];
int RH[maxn], CH[maxn], S[maxn];            // RH爲結點的行編號, CH結點的列編號, S爲每列結點數

void InitMtx() {
    int i, j, k, t;
    memset(mtx, 0, sizeof(mtx));
    for (i = 0; i < 9; i++) {
        for (j = 0; j < 9; j++) {
            t = i * 9 + j;
            if (sudoku[t] == '.') {
                for (k = 0; k < 9; k++) {
                    mtx[t*9+k][t] = 1;                      // (行,列)
                    mtx[t*9+k][81+i*9+k] = 1;               // (行,數)
                    mtx[t*9+k][162+j*9+k] = 1;              // (列,數)
                    mtx[t*9+k][243+(i/3*3+j/3)*9+k] = 1;    // (格,數)
                }
            } else {
                k = sudoku[t] - '1';
                mtx[t*9+k][t] = 1;                      // (行,列)
                mtx[t*9+k][81+i*9+k] = 1;               // (行,數)
                mtx[t*9+k][162+j*9+k] = 1;              // (列,數)
                mtx[t*9+k][243+(i/3*3+j/3)*9+k] = 1;    // (格,數)
            }
        }
    }
}

int Node(int up, int down, int left, int right) {
    U[idx] = up; D[idx] = down;
    L[idx] = left; R[idx] = right;
    D[up] = U[down] = R[left] = L[right] = idx;
    return idx++;
}

void Build() {
    int i, j, k;
    idx = maxn - 1;
    head = Node(idx, idx, idx, idx);    // 初始化列頭結點鏈表的頭結點
    idx = 0;
    for (j = 0; j < M; j++) {           // 申請M個結點爲每列的頭結點
        Node(idx, idx, L[head], head);
        CH[j] = j; S[j] = 0;
    }
    for (i = 0; i < N; i++) {
        k = -1;
        for (j = 0; j < M; j++) {
            if (!mtx[i][j]) continue;
            
            if (k == -1) {
                k = Node(U[CH[j]], CH[j], idx, idx);
                RH[k] = i; CH[k] = j; S[j]++;
            } else {
                k = Node(U[CH[j]], CH[j], k, R[k]);
                RH[k] = i; CH[k] = j; S[j]++;
            }
        }
    }
}

void Remove(int c) {
    int i, j;
    L[R[c]] = L[c];
    R[L[c]] = R[c];
    for (i = D[c]; i != c; i = D[i]) {
        for (j = R[i]; j != i; j = R[j]) {
            U[D[j]] = U[j];
            D[U[j]] = D[j];
            S[CH[j]]--;
        }
    }
}

void Resume(int c) {
    int i, j;
    R[L[c]] = c;
    L[R[c]] = c;
    for (i = U[c]; i != c; i = U[i]) {
        for (j = L[i]; j != i; j = L[j]) {
            S[CH[j]]++;
            D[U[j]] = j;
            U[D[j]] = j;
        }
    }
}

int dfs() {
    if (R[head] == head)
        return 1;
    int i, j, k, c, min = inf;
    for (j = R[head]; j != head; j = R[j]) {
        if (S[j] < min) {
            min = S[j];
            c = j;
        }
    }
    Remove(c);
    for (i = D[c]; i != c; i = D[i]) {
        k = RH[i];
        sudoku[k/9] = '1' + k % 9;
        for (j = R[i]; j != i; j = R[j])
            Remove(CH[j]);
        if (dfs()) return 1;
        for (j = L[i]; j != i; j = L[j])
            Resume(CH[j]);
    }
    Resume(c);
    return 0;
}

int main() {
    N = 729, M = 324;
    while (gets(sudoku), strcmp(sudoku, "end")) {
        InitMtx();
        Build();
        dfs();
        printf("%s\n", sudoku);
    }
    return 0;
}