NOIP2009 Target Sudoku (search)

Previously did a hdu similar Sudoku problem, the problem I did not add any optimization on the past, and then the problem is not optimized can be 50~80 points, to 80 points must be optimized

A bitwise operation can be used to save each row of each column each nine Gongne already used, for example, userow[i] = 011101111 means that line I and 1, 52 numbers are not used, then we enumerate each unknown lattice (i,j), he can not fill the number is userow[i]| USECOL[J]|USEBOX[I/3*3+J/3], in fact, this does not optimize what time, the following is the most NB optimization

We can count the known number of each line at the input, so we can sort the non-filled grid from small to large in each row, so that the search tree that the enumeration point can extend is smaller than the original.

#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #define RE (1<    &LT;9)-1) using namespace std;int sum_a[10];struct t{int x, y;        BOOL operator < (T t) const//known quantity more placed in front {if (sum_a[x] = = Sum_a[t.x]) return x < t.x;    return sum_a[x] > Sum_a[t.x];           }}unknown[100];//Save the position of the blank lattice int st[10][10];int p[10][10];//weight array void init () {for (int i = 0; i < 9; i++) {    P[0][i] = P[8][i] = p[i][0] = p[i][8] = 6;    } for (int i = 1; i < 8; i++) {p[1][i] = p[7][i] = p[i][1] = p[i][7] = 7;    } for (int i = 2; i < 7; i++) {p[2][i] = p[6][i] = p[i][2] = p[i][6] = 8;    } for (int i = 3; i < 6; i++) {p[3][i] = p[5][i] = p[i][3] = p[i][5] = 9; } P[4][4] = 10;} int ans = -1;int Cnt,sorce;int userow[10];//Each row uses the number of int usecol[10];//Each column uses the number of int usebox[10];//each nine to use the number of void Dfs (int cur        {if (cur > cnt) {ans = max (Sorce,ans);    Return  }  int x = unknown[cur].x;    int y = unknown[cur].y;    int temp = (USEROW[X]|USECOL[Y]|USEBOX[X/3*3+Y/3]); if (temp = = RE) return;//is not available for the number for (int i = 1; I <= 9; i++) if (! (        (Temp>> (i-1)) &1)) {St[x][y] = i;        USEBOX[X/3*3+Y/3] |= (1<< (i-1));        Userow[x] |= (1<< (i-1));        Usecol[y] |= (1<< (i-1));        Sorce + = St[x][y]*p[x][y];        DFS (CUR+1);        Sorce-= St[x][y]*p[x][y];        St[x][y] = 0;        USEBOX[X/3*3+Y/3] ^= (1<< (i-1));        Userow[x] ^= (1<< (i-1));    Usecol[y] ^= (1<< (i-1));    }}int Main () {//freopen ("sudoku.in", "R", stdin);    Freopen ("Sudoku.out", "w", stdout);    Init ();            for (int i = 0; i < 9; i++) {for (int j = 0; J < 9; J + +) {scanf ("%d", &st[i][j]);                if (st[i][j]! = 0) {Sorce + = St[i][j]*p[i][j];                Userow[i] |= (1<< (st[i][j]-1)); USECOL[J] |= (1<< (st[I][J]-1));                USEBOX[I/3*3+J/3] |= (1<< (st[i][j]-1));        ++sum_a[i];//Count I-line-known}}} for (int i = 0; i < 9; i++) for (int j = 0; J < 9; J + +)                {if (!st[i][j]) {++cnt;                unknown[cnt].x = i;            Unknown[cnt].y = j;    }} sort (unknown+1,unknown+cnt+1);    DFS (1); printf ("%d\n", ans);}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

NOIP2009 Target Sudoku (search)