POJ 3074 Sudoku

DLX解数独。行：81格（每格9种选择）。列：每行每列9格（各9种选择），81格（有或没有）。把已知的先加进去，并把它们标记，之后不再添加。然后把剩下的所有可能都加进去，再dfs求解，应该是唯一解。输出的时候注意一下，先求出真实位置，再求真实数值，最后直接输出字符串。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<vector>#include<string>#include<queue>#include<map>///LOOP#define REP(i, n) for(int i = 0; i < n; i++)#define FF(i, a, b) for(int i = a; i < b; i++)#define FFF(i, a, b) for(int i = a; i <= b; i++)#define FD(i, a, b) for(int i = a - 1; i >= b; i--)#define FDD(i, a, b) for(int i = a; i >= b; i--)///INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)#define RFI(n) scanf("%lf", &n)#define RFII(n, m) scanf("%lf%lf", &n, &m)#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)#define RS(s) scanf("%s", s)///OUTPUT#define PN printf("\n")#define PI(n) printf("%d\n", n)#define PIS(n) printf("%d ", n)#define PS(s) printf("%s\n", s)#define PSS(s) printf("%s ", s)#define PC(n) printf("Case %d: ", n)///OTHER#define PB(x) push_back(x)#define CLR(a, b) memset(a, b, sizeof(a))#define CPY(a, b) memcpy(a, b, sizeof(b))#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1using namespace std;typedef long long LL;typedef pair<int, int> P;const int MOD = 1e9+7;const int INFI = 1e9 * 2;const LL LINFI = 1e17;const double eps = 1e-6;const int N = 9;const int Nr = N * N * N;const int Nc = N * N * 4;const int M = Nr * Nc;const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};int U[M], D[M], L[M], R[M], X[M], Y[M], H[Nr], S[Nr], Q[Nr];bool v[M];char s[Nr];int size;void remove(int c){    L[R[c]] = L[c],R[L[c]] = R[c];    for(int i = D[c]; i != c; i = D[i])        for(int j = R[i]; j != i; j = R[j])            U[D[j]] = U[j], D[U[j]] = D[j], S[Y[j]]--;}void resume(int c){    L[R[c]] = R[L[c]] = c;    for(int i = U[c]; i != c; i = U[i])        for(int j = L[i]; j != i; j = L[j])            S[Y[U[D[j]] = D[U[j]] = j]]++;}bool dance(){    if(!R[0])    {        puts(s);        return 1;    }    int p, m = M;    for(int i = R[0]; i; i = R[i])if(m > S[i])m = S[p = i];    remove(p);    for(int i = D[p]; i != p; i = D[i])    {        s[(X[i] - 1) / 9] = (X[i] - 1) % 9 + '1';        for(int j = R[i]; j != i; j = R[j])remove(Y[j]);        if(dance())return 1;        for(int j = L[i]; j != i; j = L[j])resume(Y[j]);    }    resume(p);    return 0;}void link(int r, int c){    S[Y[size] = c]++;    D[size] = D[c];    U[size] = c;    U[D[c]] = size;    D[c] = size;    if(H[r] < 0)H[r] = L[size] = R[size] = size;    else    {        R[size] = R[H[r]];        L[R[H[r]]] = size;        L[size] = H[r];        R[H[r]] = size;    }    X[size++] = r;}int c[4];void get(int &r, int i, int j, int k){    r = (i * N + j) * N + k;    c[0] = i * N + k;    c[1] = N * N + N * j + k;    c[2] = N * N * 2 + N * ((i / 3) * 3 + j / 3) + k;    c[3] = N * N * 3 + i * N + j + 1;}void init(int r, int c){    REP(i, c + 1)L[R[i] = i + 1] = U[i] = D[i] = i;    R[c]=0;    size = c + 1;    while(r)H[r--]=-1;    CLR(S, 0);    CLR(v, 0);}int main(){    //freopen("input.txt", "r", stdin);    //freopen("output.txt", "w", stdout);    int r;    while(RS(s), s[0] != 'e')    {        init(Nr, Nc);        for(int i = 0, k = 0; i < N; i++)for(int j = 0; j < N; j++, k++)if(s[k] != '.')        {            get(r, i, j, s[k] - '0');            REP(p, 4)link(r, c[p]), v[c[p]] = 1;        }        REP(i, 9)REP(j, 9)FFF(k, 1, 9)        {            get(r, i, j, k);            if(v[c[0]] || v[c[1]] || v[c[2]] || v[c[3]])continue;            REP(p, 4)link(r, c[p]);        }        dance();    }    return 0;}