Luogu p1074 target Sudoku [Search/pruning] By cellur925

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It is obviously a search. However, there is no pruning at the beginning, and the brute force search starts from,NaturallyT has 6 points.

#include<cstdio>#include<algorithm>using namespace std;const int group[10][10]={    0,0,0,0,0,0,0,0,0,0,    0,1,1,1,2,2,2,3,3,3,    0,1,1,1,2,2,2,3,3,3,    0,1,1,1,2,2,2,3,3,3,    0,4,4,4,5,5,5,6,6,6,    0,4,4,4,5,5,5,6,6,6,    0,4,4,4,5,5,5,6,6,6,    0,7,7,7,8,8,8,9,9,9,    0,7,7,7,8,8,8,9,9,9,    0,7,7,7,8,8,8,9,9,9,};const int sco[10][10]={    0,0,0,0,0,0,0,0,0,0,    0,6,6,6,6,6,6,6,6,6,    0,6,7,7,7,7,7,7,7,6,    0,6,7,8,8,8,8,8,7,6,    0,6,7,8,9,9,9,8,7,6,    0,6,7,8,9,10,9,8,7,6,    0,6,7,8,9,9,9,8,7,6,    0,6,7,8,8,8,8,8,7,6,    0,6,7,7,7,7,7,7,7,6,    0,6,6,6,6,6,6,6,6,6,};int ans,num[50][50];bool gong[50][50],hang[50][50],lie[50][50];void review(){    int val=0;    for(int i=1;i<=9;i++)        for(int j=1;j<=9;j++)            val+=sco[i][j]*num[i][j];//  printf("%d\n",val);    ans=max(ans,val);}bool check(int x,int y,int w){    if(gong[group[x][y]][w]||hang[x][w]||lie[y][w]) return 0;    return 1;}void dfs(int x,int y){    if(x==10)    {        review();        return ;    }    int nx=x,ny=y+1;    if(ny==10) nx++,ny=1;    if(num[x][y]) dfs(nx,ny);    else    {        for(int i=1;i<=9;i++)        {            if(!check(x,y,i)) continue;            gong[group[x][y]][i]=1;            hang[x][i]=1;            lie[y][i]=1;            num[x][y]=i;            dfs(nx,ny);            gong[group[x][y]][i]=0;            hang[x][i]=0;            lie[y][i]=0;            num[x][y]=0;        }    }}int main(){    for(int i=1;i<=9;i++)        for(int j=1;j<=9;j++)        {            int x=0;            scanf("%d",&x);            if(!x) continue;            num[i][j]=x;            hang[i][x]=1;            lie[j][x]=1;            gong[group[i][j]][x]=1;        }    dfs(1,1);       printf("%d",ans==0 ? -1 : ans);    return 0;}

Consider pruning. From the perspective of human intelligenceAlthough I have never playedWe must start from a region with more options, because fewer decisions are available for us. So here we can also use this idea for reference. Every time we make statistics on the number of columns in each row, we have already filled in the number of records and obtained an optimal coordinate from which we can start searching. This algorithm ensures that we do not enter a number for each search, which makes the complexity much better.

# Include <cstdio> # include <algorithm> using namespace STD; const int group [10] [10] =,, 5, 5, 5, 6, 6, 0, 4, 4, 5, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 7, 8, 8, 9, 9, 7, 7, 8, 8, 8, 9, 9,}; const int SCO [10] [10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,, 6, 7, 8, 9, 9, 8, 7, 6, 9, 10, 9, 8, 7, 6, 7, 8, 9, 9, 8, 7, 6, 7, 8, 8, 8, 8, 8,, int CN, ANS, num [50] [50]; int cnt_hang [50], cnt_lie [50]; bool Gong [50] [50], hang [50] [50], lie [50] [50]; void Review () {int val = 0; For (INT I = 1; I <= 9; I ++) for (Int J = 1; j <= 9; j ++) val + = SCO [I] [J] * num [I] [J]; // printf ("% d \ n", Val); ans = max (ANS, val);} bool check (int x, int y, int W) {If (Gong [GRO Up [x] [Y] [W] | hang [x] [W] | lie [y] [W]) return 0; return 1 ;} void DFS (int x, int y, int CNT) {If (CNT = 81) {Review (); Return ;}for (INT I = 1; I <= 9; I ++) {If (! Check (x, y, I) continue; Gong [Group [x] [Y] [I] = 1; hang [x] [I] = 1; lie [y] [I] = 1; num [x] [Y] = I; cnt_hang [x] ++; cnt_lie [y] ++; int qwq =-1, qaq =-1, BX = 0, by = 0; For (Int J = 1; j <= 9; j ++) if (cnt_hang [J]> qwq & cnt_hang [J] <9) qwq = cnt_hang [J], BX = J; For (Int J = 1; j <= 9; j ++) if (cnt_lie [J]> Qaq &&(! Num [BX] [J]) Qaq = cnt_lie [J], by = J; DFS (BX, by, CNT + 1 ); gong [Group [x] [y] [I] = 0; hang [x] [I] = 0; lie [y] [I] = 0; num [x] [Y] = 0; cnt_hang [x] --; cnt_lie [y] -- ;}} int main () {for (INT I = 1; I <= 9; I ++) for (Int J = 1; j <= 9; j ++) {int x = 0; scanf ("% d ", & X); If (! X) continue; num [I] [J] = x; cnt_hang [I] ++; cnt_lie [J] ++; hang [I] [x] = 1; lie [J] [x] = 1; Gong [Group [I] [J] [x] = 1; CN ++;} int qwq =-1, qaq =-1, BX = 0, by = 0; For (INT I = 1; I <= 9; I ++) if (cnt_hang [I]> qwq & cnt_hang [I] <9) qwq = cnt_hang [I], BX = I; for (INT I = 1; I <= 9; I ++) if (cnt_lie [I]> Qaq &&(! Num [BX] [I]) // you can find a coordinate Qaq = cnt_lie [I], by = I; DFS (BX, by, CN) that is not filled in ); printf ("% d", ANS = 0? -1: ANS); Return 0 ;}

Warning

I started writing my own brute-force statement twice:

① No input/output (???) Sure enough, t is hopeless.

② Because I'm sure the palace and the value are both played out by the array table, and started to open the array very large, \ (50*50 \), however, the part of the table we typed is not a separate row and will be understood as a continuous segment after compilation. Therefore, we need to limit the array size so that we can just fill in the number.

Luogu p1074 target Sudoku [Search/pruning] By cellur925