solution

• 找出满足①像素值唯一&②和相邻像素点色差大于tol 的像素点个数
  • 若唯一，则输出该像素点列、行、像素值；
  • 若不唯一，则输出"Not Unique"
  • 若无，则输出"Not Exist"
• 题干和输出中，都是先列再行

#include<iostream>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1010;
ll a[maxn][maxn];
int main(){
	int m, n, cnt = 0, flag, x, y;
	ll tol, color;
	map<ll, int> mp;
	scanf("%d%d%lld", &m, &n, &tol);
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			scanf("%lld", &a[i][j]);
			mp[a[i][j]]++;
		}
	}
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			flag = 0;
			if(mp[a[i][j]] == 1){
				for(int k = i - 1; k <= i + 1 && !flag; k++){
					for(int l = j - 1; l <= j + 1 && !flag; l++){
						if(k < 1 || k > n || l < 1 || l > m || (k == i && l == j)) continue;
						if(abs(a[i][j] - a[k][l]) <= tol) flag = 1;
					}
				}
				if(!flag) {
					cnt++;
					x = j;
					y = i;
					color = a[i][j];
				}
			}
		}
	}
	if(!cnt) printf("Not Exist");
	else if(cnt > 1) printf("Not Unique");
	else printf("(%d, %d): %lld", x, y, color);
	return 0;
}