POJ 3076 Sudoku (DLX解数独)

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题目大意：
解一个16*16形式的数独
要求每行每列每个16宫格都包含A~P
大致思路:
和POJ 3074一个道理...继续练手
刚开始数组开小调了一段时间...
代码如下：
Result : Accepted Memory : 500 KB Time : 3391 ms
/* * Author: Gatevin * Created Time:  2015/10/2 8:52:27 * File Name: Sakura_Chiyo.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;/* * DLX解经典数独问题 * 考虑到每行都要有1~16 * 用第1~16*16列表示第i行是否有数字j(第(i - 1)*9 + j列的状态） * 由于每列都要有1~16 * 用第16*16 + 1 ~  16*16*2列表示第i列是否有数字j (第(i - 1)*16 + j + 16*16列的状态) * 考虑到每个九宫格都要有1~16 * 用第16*16+1~16*16*2列表示第i个九宫格是否有数字j (第(i - 1)*16 + j + 16*16列的状态) * 但是仅仅依靠这些是不够的 * 还有一个限制条件每个格子只能有1个数字 * 所以用第16*16*2+1~16*16*3列表示格子中是否填了数 * 那么对于给定的数字, 就在对应的列上写4个1 * 如果是.那么就添加16行, 分别对应哪个位子填1~16的情况 * 那么最坏情况下有16*16*16行可选 */char in[800];struct DLX{#define maxn 16*16*16 + 4#define maxm 16*16*4 + 4#define maxnode 16*16*16*4 + 100    int n, m, size;    int U[maxnode], D[maxnode], R[maxnode], L[maxnode], Row[maxnode], Col[maxnode];    int H[maxn], S[maxm];    int ansd, ans[maxn];    struct State    {        int x, y, val;        State(int _x, int _y, int _val)        {            x = _x, y = _y, val = _val;        }        State(){}    };    State s[maxn];        void init(int _n, int _m)    {        n = _n;        m = _m;        for(int i = 0; i <= m; i++)        {            S[i] = 0;            U[i] = D[i] = i;            L[i] = i - 1;            R[i] = i + 1;        }        R[m] = 0; L[0] = m;        size = m;        for(int i = 1; i <= n; i++) H[i] = -1;    }    void Link(int r, int c)    {        ++S[Col[++size] = c];        Row[size] = r;        D[size] = D[c];        U[D[c]] = size;        U[size] = c;        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;        }    }    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[Col[j]];            }    }    void resume(int c)    {        for(int i = U[c]; i != c; i = U[i])            for(int j = L[i]; j != i; j = L[j])                ++S[Col[U[D[j]] = D[U[j]] = j]];        L[R[c]] = R[L[c]] = c;    }    bool Dance(int dep)    {        if(R[0] == 0)        {            ansd = dep;            return true;        }        int c = R[0];        for(int i = R[0]; i != 0; i = R[i])            if(S[i] < S[c])                c = i;        remove(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]) remove(Col[j]);            if(Dance(dep + 1)) return true;            for(int j = L[i]; j != i; j = L[j]) resume(Col[j]);        }        resume(c);        return false;    }    void solve()    {        int tn = 0, tm = 16*16*4;        int len = strlen(in);        for(int i = 0; i < len; i++)            if(in[i] == '-') tn += 16;            else tn++;        init(tn, tm);        tn = 0;        for(int i = 0; i < 16; i++)            for(int j = 0; j < 16; j++)            {                char c = in[i*16 + j];                if(c == '-')                {                    for(int k = 1; k <= 16; k++)                    {                        s[++tn] = State(i, j, k);                        Link(tn, i*16 + k);                        Link(tn, j*16 + k + 16*16);                        Link(tn, (i/4 * 4 + j / 4)*16 + k + 16*16*2);                        Link(tn, i*16 + j + 1 + 16*16*3);                    }                }                else                {                    s[++tn] = State(i, j, c - 'A' + 1);                    Link(tn, i*16 + c - 'A' + 1);                    Link(tn, j*16 + c - 'A' + 1 + 16*16);                    Link(tn, (i/4 * 4 + j / 4)*16 + c - 'A' + 1 + 16*16*2);                    Link(tn, i*16 + j + 1 + 16*16*3);                }            }        if(!Dance(0)) puts("No answer");        else        {            for(int i = ansd - 1; i >= 0; i--)            {                int sel = ans[i];                in[s[sel].x*16 + s[sel].y] = s[sel].val + 'A' - 1 ;            }            for(int i = 0; i < 16*16; i++)            {                putchar(in[i]);                if(i % 16 == 15) putchar('\n');            }        }    }};DLX dlx;int main(){    bool fir = 1;//每组数据间要空行不然会PE...    while(scanf("%s", in) != EOF)    {        //if(strcmp(in, "end") == 0) break;        if(!fir) putchar('\n');        fir = 0;        for(int i = 1; i < 16; i++)            scanf("%s", in + i*16);        dlx.solve();    }    return 0;}
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