SJTU 机试 最小面积子矩阵 压缩+双指针

链接：https://www.nowcoder.com/questionTerminal/8ef506fbab2742809564e1a288358554
来源：牛客网

一个N*M的矩阵，找出这个矩阵中所有元素的和不小于K的面积最小的子矩阵（矩阵中元素个数为矩阵面积）

输入描述:

每个案例第一行三个正整数N,M<=100，表示矩阵大小，和一个整数K
接下来N行，每行M个数，表示矩阵每个元素的值

输出描述:

输出最小面积的值。如果出现任意矩阵的和都小于K，直接输出-1。

示例1

输入

4 4 10
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

输出

1
和NOIp 最大加权矩阵一个套路；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int n, m, K;
int a[200][200];
bool fg;
int pre[maxn];
int dp[maxn];
int minn = inf;
void sol() {
	for (int i = 1; i <= n; i++) {
		ms(dp);
		for (int j = i; j <= n; j++) {
			ms(pre);
			for (int k = 1; k <= m; k++)
				dp[k] += a[j][k];
			for (int k = 1; k <= m; k++)pre[k] = pre[k - 1] + dp[k];
			int l = 1, r = 1;
			while (1) {
				if (r > m)break;
				while (pre[r] - pre[l - 1] < K&&r <= m)r++;
				if (pre[r] - pre[l - 1] >= K&&r<=m) {
					minn = min(minn, (j - i + 1)*(r - l + 1));
					fg = 1;
				}
				while (pre[r] - pre[l - 1] >= K && l <= r) {
					minn = min(minn, (r - l + 1)*(j - i + 1));
					l++;
					fg = 1;
				}
			}
		}

	}
}
int main()
{
	//ios::sync_with_stdio(0);
	rdint(n); rdint(m); rdint(K);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++)rdint(a[i][j]);
	}
	sol();
	if (fg == 0)cout << -1 << endl;
	else {
		cout << minn << endl;
	}
	return 0;
}