传送门：点击打开链接
题意：类似数独，不过还有另外一种限制，从左往右看，从右往左看，从上往下看，从下往上看，看到的塔数应该是固定的。
思路：暴搜。好久没暴搜了，呼呼。。
#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int>PII;const int MX = 1e2 + 5;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;char s[MX][MX];
int vis[2][MX][MX], r, n;
struct Point {int x, y;
} P[MX];bool left(int i, int tot) {int ret = 0, now = 0, cnt = 0;for(int j = 1; j <= n; j++) {if(s[i][j] > now) now = s[i][j], ret++;if(s[i][j]) cnt++;}if(cnt == n) return ret == tot;return ret <= tot;
}
bool right(int i, int tot) {int ret = 0, now = 0, cnt = 0;for(int j = n; j >= 1; j--) {if(s[i][j] > now) now = s[i][j], ret++;if(s[i][j]) cnt++;}if(cnt == n) return ret == tot;return true;
}
bool top(int j, int tot) {int ret = 0, now = 0, cnt = 0;for(int i = 1; i <= n; i++) {if(s[i][j] > now) now = s[i][j], ret++;if(s[i][j]) cnt++;}if(cnt == n) return ret == tot;return ret <= tot;
}
bool foot(int j, int tot) {int ret = 0, now = 0, cnt = 0;for(int i = n; i >= 1; i--) {if(s[i][j] > now) now = s[i][j], ret++;if(s[i][j]) cnt++;}if(cnt == n) return ret == tot;return true;
}
bool check(int i, int j, int t) {if(vis[0][i][t]) return false;if(vis[1][j][t]) return false;s[i][j] = t;if(s[i][0] && !left(i, s[i][0])) return s[i][j] = 0;if(s[i][n + 1] && !right(i, s[i][n + 1])) return s[i][j] = 0;if(s[0][j] && !top(j, s[0][j])) return s[i][j] = 0;if(s[n + 1][j] && !foot(j, s[n + 1][j])) return s[i][j] = 0;vis[0][i][t] = vis[1][j][t] = 1;return true;
}
bool check_all() {for(int i = 0; i <= n + 1; i++) {for(int j = 0; j <= n + 1; j++) {s[i][j] = isdigit(s[i][j]) ? s[i][j] - '0' : 0;if(1 <= i && i <= n && 1 <= j && j <= n) {int t = s[i][j];if(t) {if(vis[0][i][t]) return false;if(vis[1][j][t]) return false;vis[0][i][t] = vis[1][j][t] = 1;} else P[r].x = i, P[r].y = j, r++;}}}for(int i = 1; i <= n; i++) {if(s[i][0] && !left(i, s[i][0])) return false;if(s[i][n + 1] && !right(i, s[i][n + 1])) return false;if(s[0][i] && !top(i, s[0][i])) return false;if(s[n + 1][i] && !foot(i, s[n + 1][i])) return false;}return true;
}
bool DFS(int c) {if(c == r) return true;for(int x = 1; x <= n; x++) {if(!check(P[c].x, P[c].y, x)) continue;if(DFS(c + 1)) return true;vis[0][P[c].x][x] = vis[1][P[c].y][x] = s[P[c].x][P[c].y] = 0;}return false;
}
int main() {int T; //FIN;scanf("%d", &T);while(T--) {r = 0;memset(vis, 0, sizeof(vis));scanf("%d", &n);for(int i = 0; i <= n + 1; i++) {scanf("%s", s[i]);}if(!check_all() || !DFS(0)) printf("no\n\n");else {for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) printf("%d", (int)s[i][j]);printf("\n");}printf("\n");}}return 0;
}