USACO17OPEN Modern Art

题目传送门

很妙的差分题目

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分别找到画布上每种颜色的最上、最下、最左和最右，然后进行差分，求出每个位置至少涂了几次。如果涂的次数> 1，则肯定不是先涂的
有一种特殊情况，如果涂的次数== 1，但只有一种颜色，且n != 1，则一定不是先涂的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL read() {
	LL k = 0, f = 1; char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9')
		k = k * 10 + c - 48, c = getchar();
	return k * f;
}
int u[1000010], d[1000010], l[1000010], r[1000010], cnt;
int col[1010][1010], x[1010][1010], ans;
bool vis[1000010];
int main() {
	int n = read();
	memset(u, 127, sizeof(u)); memset(l, 127, sizeof(l));
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= n; ++j) {
			int a = read(); x[i][j] = a;
			if(!d[a] && a) ++cnt;
			u[a] = min(u[a], i); d[a] = max(d[a], i);
			l[a] = min(l[a], j); r[a] = max(r[a], j);
		}
	for(int i = 1; i <= n*n; ++i) {
		if(!d[i]) continue;
		col[u[i]][l[i]] += 1;
		col[u[i]][r[i]+1] -= 1;
		col[d[i]+1][l[i]] -= 1;
		col[d[i]+1][r[i]+1] += 1;
	}
	for(int i = 1; i <= n; ++i) {
		for(int j = 1; j <= n; ++j) {
			col[i][j] += col[i][j-1] + col[i-1][j] - col[i-1][j-1];
			if(x[i][j] && col[i][j] > 1)
				vis[x[i][j]] = 1;
		}
	}
	for(int i = 1; i <= n*n; ++i)
		if(!vis[i]) ++ans;
	if(cnt == 1 && n > 1) --ans;
	cout << ans << endl;
	
	return 0;
}