这份代码应该可以被卡到 O(n4),但是我不会造 hack,求助万能的谷民!
#include <bits/stdc++.h>
using namespace std;
int t, n, m, c[510][510];
bool f(int x, int y) {
int a = c[x][y], b = c[x][y + 1],
c1 = c[x + 1][y], d = c[x + 1][y + 1];
if (!a && !b && !c1 && !d) return 0;
int bas = 0;
if (a) bas = a;
if (b) bas = b;
if (c1) bas = c1;
if (d) bas = d;
if ((!a || a == bas)
&& (!b || b == bas)
&& (!c1 || c1 == bas)
&& (!d || d == bas)) {
return 1;
}
return 0;
}
int main() {
// freopen("TestData020.in", "r", stdin);
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t--) {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cin >> c[i][j];
}
}
int p = 0;
while (1) {
bool flag = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
if (c[i][j])
flag = 1;
if (!flag) {
p = 1;
break;
}
flag = 0;
for (int i = 1; i < n; i++) {
for (int j = 1; j < m; j++) {
if (f(i, j)) {
c[i][j] =
c[i][j + 1] =
c[i + 1][j] =
c[i + 1][j + 1] = 0;
flag = 1;
}
}
}
if (!flag) break;
}
if (p) cout << "Yes\n";
else cout << "No\n";
}
return 0;
}