POJ 3076 Sudoku

3076

思路：

dfs + 剪枝

首先，如果这个位置只能填一种字母，那就直接填

其次，如果对于每一种字母，如果某一列或者某一行或者某一块只能填它，那就填它

然后，对于某个位置如果不能填字母了，或者某种字母在一行一列或一块中出向了两次以上，说明当前方案不成立

最后贪心地从可选情况少的往下搜

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 17;
int mp[N][N];
int st[N][N];
int sum = 0;
char s[N][N+5];
void add(int x, int y, int t) {
    mp[x][y] = t;
    sum++;
    for (int i = 1; i <= 16; i++) {
        st[i][y] |= 1<<t-1;
        st[x][i] |= 1<<t-1;
    }
    int xx = (x+3)/4, yy = (y+3)/4;
    for (int i = (xx-1)*4 + 1; i <= xx*4; i++) {
        for (int j = (yy-1)*4 + 1; j <= yy*4; j++) {
            st[i][j] |= 1<<t-1;
        }
    }
}
void print() {
    for (int i = 1; i <= 16; i++) {
        for (int j = 1; j <= 16; j++) {
            putchar(mp[i][j]-1+'A');
        }
        puts("");
    }
    puts("");
}
bool dfs() {
    if(sum == 256) {
        print();
        return true;
    }

    for (int i = 1; i <= 16; i++) {
        for (int j = 1; j <= 16; j++) {
            if(!mp[i][j]) {
                int cnt = 0, t = 0;
                for (int k = 1; k <= 16; k++) {
                    if((st[i][j] & (1<<k-1)) == 0) {
                        cnt++;
                        t = k;
                        if(cnt == 2) break;
                    }
                }
                if(!cnt) return false;
                if(cnt == 1) add(i, j, t);
            }
        }
    }

    for (int i = 1; i <= 16; i++) {
        for (int k = 1; k <= 16; k++) {
            int cnt1 = 0, cnt2 = 0, y;
            for (int j = 1; j <= 16; j++) {
                if(mp[i][j] == k) cnt1++;
                if(cnt1 == 2) return false;
                if(!mp[i][j] && (st[i][j] & (1<<k-1)) == 0) cnt2++, y = j;
            }
            if(!cnt1 && !cnt2) return false;
            if(!cnt1 && cnt2 == 1) add(i, y, k);
        }
    }

    for (int j = 1; j <= 16; j++) {
        for (int k = 1; k <= 16; k++) {
            int cnt1 = 0, cnt2 = 0, x;
            for (int i = 1; i <= 16; i++) {
                if(mp[i][j] == k) cnt1++;
                if(cnt1 == 2) return false;
                if(!mp[i][j] && (st[i][j] & (1<<k-1)) == 0) cnt2++, x = i;
            }
            if(!cnt1 && !cnt2) return false;
            if(!cnt1 && cnt2 == 1) add(x, j, k);
        }
    }

    for (int i = 1; i <= 16; i++) {
        int x = (i+3)/4, y = i - (x-1)*4;
        for (int k = 1; k <= 16; k++) {
            int cnt1 = 0, cnt2 = 0, xx, yy;
            for (int ii = (x-1)*4+1; ii <= x*4; ii++) {
                for (int jj = (y-1)*4+1; jj <= y*4; jj++) {
                    if(mp[ii][jj] == k) cnt1++;
                    if(cnt1 == 2) return false;
                    if(!mp[ii][jj] && (st[ii][jj] & (1<<k-1)) == 0) cnt2++, xx = ii, yy = jj;
                }
            }
            if(!cnt1 && !cnt2) return false;
            if(!cnt1 && cnt2 == 1) add(xx, yy, k);
        }
    }
    if(sum == 256) {
        print();
        return true;
    }

    int mn = N, x, y;
    for (int i = 1; i <= 16; i++) {
        for (int j = 1; j <= 16; j++) {
            if(!mp[i][j]) {
                int cnt = 0;
                for (int k = 1; k <= 16; k++) {
                    if((st[i][j] & (1<<k-1)) == 0) {
                        cnt++;
                        if(cnt >= mn) break;
                    }
                }
                if(cnt < mn) {
                    mn = cnt;
                    x = i;
                    y = j;
                }
            }
        }
    }
    int tst[N][N], tmp[N][N];
    memcpy(tst, st, sizeof(st));
    memcpy(tmp, mp, sizeof(mp));
    int tsum = sum;

    for (int k = 1; k <= 16; k++) {
        if((st[x][y] & (1<<k-1)) == 0) {
            add(x, y, k);
            bool f = dfs();
            if(!f) {
                memcpy(st, tst, sizeof(tst));
                memcpy(mp, tmp, sizeof(tmp));
                sum = tsum;
            }
            else return true;
        }
    }
    return false;
}
int main() {
    while(scanf("%s", s[1]+1) != EOF){
        for (int i = 2; i <= 16; i++) {
            scanf("%s", s[i]+1);
        }
        sum = 0;
        mem(mp, 0);
        mem(st, 0);
        for (int i = 1; i <= 16; i++) {
            for (int j = 1; j <= 16; j++) {
                if(isalpha(s[i][j])) add(i, j, s[i][j] - 'A' + 1);
            }
        }
        dfs();
    }
    return 0;
}
/*
--A----C-----O-I
-J--A-B-P-CGF-H-
--D--F-I-E----P-
-G-EL-H----M-J--
----E----C--G---
-I--K-GA-B---E-J
D-GP--J-F----A--
-E---C-B--DP--O-
E--F-M--D--L-K-A
-C--------O-I-L-
H-P-C--F-A--B---
---G-OD---J----H
K---J----H-A-P-L
--B--P--E--K--A-
-H--B--K--FI-C--
--F---C--D--H-N-
*/