poj3076Sudoku(我大DLX威武！16*16数独秒解！！)

来源：互联网发布：tomcat centos 部署编辑：程序博客网时间：2024/06/08 02:29
->题目请戳这里<-
题目大意：16*16数独。
题目分析：建模同这篇9*9数独，就不多废话了，详情请见代码：
#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 16 * 16 * 16 * 16 * 16 * 4 + 5;//16 * 16 * 16行16 * 16 * 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[20][20];void init(){    memset(s,0,sizeof(s));//重要初始化，否则TLE！！    head = 0;    int i;    int c = 16 * 16 * 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)    {        int t = 0;        sort(ans,ans + 16 * 16);        for(i = 0;i < 16;i ++)        {            for(j = 0;j < 16;j ++)            {                printf("%c",ans[t ++] - 16 * (i * 16 + j) + 64);            }            printf("\n");        }        //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(){    int i,j;    while(scanf("%s",str[0]) != EOF)    {        for(i = 1;i < 16;i ++)            scanf("%s",str[i]);        init();        //system("pause");        for(i = 0;i < 16;i ++)        {            for(j = 0;j < 16;j ++)            {                int base = (i * 16 + j) * 16;                if(str[i][j] == '-')                {                    for(int k = 1;k <= 16;k ++)//依次填充1-16                    {                        insert(base + k,i * 16 + k);//第i行填k                        insert(base + k,256 + 16 * j + k);//第j列填k                        insert(base + k,256 + 256 + (4 * (i / 4) + j / 4) * 16 + k);                        insert(base + k,256 + 256 + 256 + i * 16 + j + 1);                    }                }                else                {                    insert(base + str[i][j] - 64,i * 16 + str[i][j] - 64);                    insert(base + str[i][j] - 64,256 + 16 * j + str[i][j] - 64);                    insert(base + str[i][j] - 64,256 + 256 + (4 * (i / 4) + j / 4) * 16 + str[i][j] - 64);                    insert(base + str[i][j] - 64,256 + 256 + 256 + i * 16 + j + 1);                }            }        }        dfs(0);    }    return 0;}