POJ 3076 Sudoku 精确覆盖问题DLX

题目：

http://poj.org/problem?id=3076

题意：

16宫格，用大写字母A~P填充，输出一种方案

思路

跳舞链。注意建列约束的时候，按照存在性、行、列、块的顺序比行、列、块、存在性的顺序建约束快了差不多10倍，很迷。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int X = 300000 + 10, N = 5000 + 10, M = 1200 + 10, INF = 0x3f3f3f3f;

struct DLX
{
    int U[X], D[X], L[X], R[X], row[X], col[X];
    int H[N], S[M];
    int head, sz, tot, n, m, ans[N];
    void init(int _n, int _m)
    {
        n = _n, m = _m;
        for(int i = 0; i <= m; i++)
            L[i] = i-1, R[i] = i+1, U[i] = D[i] = i, S[i] = 0;
        head = 0, tot = 0, sz = m;
        L[head] = m, R[m] = head;
        for(int i = 1; i <= n; i++) H[i] = -1;
    }
    void link(int r, int c)
    {
        ++S[col[++sz]=c];
        row[sz] = r;
        D[sz] = D[c], U[D[c]] = sz;
        U[sz] = c, D[c] = sz;
        if(H[r] < 0) H[r] = L[sz] = R[sz] = sz;
        else R[sz] = R[H[r]], L[R[H[r]]] = sz, L[sz] = H[r], R[H[r]] = sz;
    }
    void del(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])
                D[U[j]] = D[j], U[D[j]] = U[j], --S[col[j]];
    }
    void recover(int c)
    {
        for(int i = U[c]; i != c; i = U[i])
            for(int j = L[i]; j != i; j = L[j])
                D[U[j]] = U[D[j]] = j, ++S[col[j]];
        R[L[c]] = L[R[c]] = c;
    }
    bool dance(int dep)
    {
        if(R[head] == head)
        {
            tot = dep-1; return true;
        }
        int c = R[head];
        for(int i = R[head]; i != head; i = R[i])
            if(S[i] < S[c]) c = i;
        del(c);
        for(int i = D[c]; i != c; i = D[i])
        {
            ans[dep] = row[i];
            for(int j = R[i]; j != i; j = R[j]) del(col[j]);
            if(dance(dep + 1)) return true;
            for(int j = L[i]; j != i; j = L[j]) recover(col[j]);
        }
        recover(c);
        return false;
    }
}dlx;
int encode(int a, int b, int c, int len)
{
    return a*len*len + b*len + c;
}
void calc(int x, int y, int k, int len)
{
    int r = ((x-1) * len + (y-1)) * len + k;
    dlx.link(r, encode(0, x-1, y, len));
    dlx.link(r, encode(1, x-1, k, len));
    dlx.link(r, encode(2, y-1, k, len));

    int base = 4;
    int block = (x-1) / base * base + (y-1) / base + 1;
    dlx.link(r, encode(3, block-1, k, len));
//    dlx.link(r, (x-1) * len + y);
//    dlx.link(r, len*len + (x-1) * len + k);
//    dlx.link(r, len*len*2 + (y-1) * len + k);
//
//    int base = 4;
//    int block = (x-1) / base * base + (y-1) / base + 1;
//    dlx.link(r, len*len*3 + (block-1) * len + k);
}
//有两个不同的输出函数，用哪一个都可以，之前用的注释掉的这个，这次用另外一个
//void print(int len)
//{
//    sort(dlx.ans + 1, dlx.ans + 1 + dlx.tot);
//    for(int i = 1; i <= len; i++)
//    {
//        for(int j = 1; j <= len; j++)
//            printf("%c", dlx.ans[(i-1)*len+j] - ((i-1)*len + (j-1)) * len - 1 + 'A');
//        printf("\n");
//    }
//}
void recode(int val, int &a, int &b, int &c)
{
    val--;
    c = val % 16; val /= 16;
    b = val % 16; val /= 16;
    a = val;
}
void print(int len, char str[][20])
{
    int x, y, k;
    for(int i = 1; i <= dlx.tot; i++)
    {
        recode(dlx.ans[i], x, y, k);
        str[x+1][y+1] = 'A' + k;
    }
    for(int i = 1; i <= len; i++) printf("%s\n", str[i] + 1);
}
int main()
{
    int len = 16, cas = 0;
    char str[20][20];
    while(~ scanf("%s", str[1]+1))
    {
        for(int i = 2; i <= len; i++) scanf("%s", str[i]+1);
        if(cas++ != 0) printf("\n"); 
        dlx.init(len * len * len, len * len * 4);
        for(int i = 1; i <= len; i++)
            for(int j = 1; j <= len; j++)
                if(str[i][j] == '-')
                    for(int k = 1; k <= len; k++) calc(i, j, k, len);
                else calc(i, j, str[i][j]-'A'+1, len);
        dlx.dance(1);
        //print(len);
        print(len, str);
    }
    return 0;
}