Sudoku (剪枝+状态压缩+预处理）

【题目描述】

	In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For example,

.	2	7	3	8	.	.	1	.
.	1	.	.	.	6	7	3	5
.	.	.	.	.	.	.	2	9
3	.	5	6	9	2	.	8	.
.	.	.	.	.	.	.	.	.
.	6	.	1	7	4	5	.	3
6	4	.	.	.	.	.	.	.
9	5	1	8	.	.	.	7	.
.	8	.	.	6	5	3	4	.

	Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.	

【题目链接】

Sudoku

【算法】

剪枝+状态压缩常数优化。好难啊。。。。

【代码】

#include <stdio.h>
using namespace std;
char G[10][10];
int row[10],col[10],grid[10],rec[512],num[512];
inline int g(int x,int y) {
    return (x/3)*3+y/3;
}
inline void flip(int x,int y,int to) {
    row[x]^=1<<to;
    col[y]^=1<<to;
    grid[g(x,y)]^=1<<to;
}
bool dfs(int cur) {
    if(cur==0) return 1;
    int minn=10,x,y;
    for(int i=0;i<9;i++) {
        for(int j=0;j<9;j++) {
            if(G[i][j]=='.') {
                int val=row[i]&col[j]&grid[g(i,j)];
                if(!val) return 0;
                if(rec[val]<minn) {
                    minn=rec[val],x=i,y=j;
                }
            }
        }
    }
    int val=row[x]&col[y]&grid[g(x,y)];
    for(;val;val-=val&-val) {
        int to=num[val&-val];
        G[x][y]=to+'1';
        flip(x,y,to);
        if(dfs(cur-1)) return 1;
        flip(x,y,to);
        G[x][y]='.';
    }
    return 0;
}
int main() {
    for(int i=0;i<1<<9;i++) {
        for(int j=i;j;j-=j&-j)
            rec[i]++;
    }
    for(int i=0;i<9;i++) {
        num[1<<i]=i;
    }
    char s[100];
    while(~scanf("%s",s)&&s[0]!='e') {
        for(int i=0;i<9;i++) row[i]=col[i]=grid[i]=(1<<9)-1;
        int tot=0;
        for(int i=0;i<9;i++) {
            for(int j=0;j<9;j++) {
                G[i][j]=s[i*9+j];
                if(G[i][j]!='.') flip(i,j,G[i][j]-'1');
                else ++tot;
            }
        }
        dfs(tot);
        for(int i=0;i<9;i++)
            for(int j=0;j<9;j++)
                s[i*9+j]=G[i][j];
        puts(s);
    }
    return 0;
}