poj3074Sudoku(DLX解9*9数独)

->题目请戳这里<-

题目大意：填9*9数独，就不多废话了。

题目分析：以前写数独是用搜索做的，略带暴力枚举，代码量比较大，效率也不高，最近学dancing links，用来解决精确覆盖问题，效率蛮高，而且代码非常优雅。关于数独转化成精确覆盖问题，这篇论文讲的很详细，就不在这里班门弄斧了，详情请见代码：

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 9 * 9 * 9 * 9 * 9 * 4 + 5;//9 * 9 * 9行9 * 9 * 4列int s[N],h[N],u[N],d[N],l[N],r[N],col[N],row[N],ans[N];int head,num;char str[200];void init(){    memset(s,0,sizeof(s));//记得初始化，否则TLE！！    head = 0;    int i;    int c = 9 * 9 * 4;    for(i = 0;i <= c;i ++)    {        u[i] = d[i] = i;        l[i] = (i - 1 + c + 1) % (c + 1);        r[i] = (i + 1) % (c + 1);    }    num = c + 1;    memset(h,0,sizeof(h));}void insert(int i,int j){    if(h[i])    {        r[num] = h[i];        l[num] = l[h[i]];        l[r[num]] = num;        r[l[num]] = num;    }    else    {        h[i] = num;        l[num] = r[num] = num;    }    s[j] ++;    u[num] = u[j];    d[num] = j;    d[u[j]] = num;    u[j] = num;    col[num] = j;    row[num] = i;    num ++;}void del(int c){    l[r[c]] = l[c];    r[l[c]] = r[c];    int i,j;    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[col[j]] --;        }    }}void recover(int c){    int i,j;    for(i = u[c];i != c;i = u[i])    {        for(j = l[i];j != i;j = l[j])        {            s[col[j]] ++;            u[d[j]] = d[u[j]] = j;        }    }    r[l[c]] = l[r[c]] = c;}bool dfs(int k){    int i,j;    if(r[head] == head)    {        sort(ans,ans + 81);        for(i = 0;i < 81;i ++)        {            printf("%d",ans[i] - i * 9);        }        printf("\n");        return true;    }    int mn = N;    int c;    for(i = r[head];i != head;i = r[i])    {        if(s[i] < mn)        {            mn = s[i];            c = i;        }    }    del(c);    for(i = d[c];i != c;i = d[i])    {        ans[k] = row[i];        for(j = r[i];j != i;j = r[j])            del(col[j]);        if(dfs(k + 1))            return true;        for(j = l[i];j != i;j = l[j])            recover(col[j]);    }    recover(c);    return false;}int main(){    while(scanf("%s",str) != EOF)    {        if(str[0] == 'e')            break;        init();        //system("pause");        for(int i = 0;i < strlen(str);i ++)        {            int rr = i / 9;            int cc = i - rr * 9;            int base = (rr * 9 + cc) * 9;            if(str[i] == '.')            {                for(int k = 1;k <= 9;k ++)//依次填充1-9                {                    insert(base + k,rr * 9 + k);//第rr行填k                    insert(base + k,81 + 9 * cc + k);//第cc列填k                    insert(base + k,81 + 81 + (3 * (rr / 3) + cc / 3) * 9 + k);                    insert(base + k,81 + 81 + 81 + rr * 9 + cc + 1);                }            }            else            {                insert(base + str[i] - '0',rr * 9 + str[i] - '0');                insert(base + str[i] - '0',81 + 9 * cc + str[i] - '0');                insert(base + str[i] - '0',81 + 81 + (3 * (rr / 3) + cc / 3) * 9 + str[i] - '0');                insert(base + str[i] - '0',81 + 81 + 81 + rr * 9 + cc + 1);            }        }        dfs(0);    }    return 0;}//7568K375MS