AtCoder - 2565 枚举+贪心

最新推荐文章于 2022-09-06 01:40:46 发布

转载最新推荐文章于 2022-09-06 01:40:46 发布 · 123 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zxyqzy/p/10319198.html

文章标签：

#c/c++

探讨了如何将一块矩形巧克力条均匀分成三块，使各块面积之差最小，提出了一种通过枚举切割位置并贪心选择剩余部分中间区域的算法。

There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle.

Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_max - S_min, where S_max is the area (the number of blocks contained) of the largest piece, and S_min is the area of the smallest piece. Find the minimum possible value of S_max−S_min.

Constraints

2≤H,W≤10⁵

Input

Input is given from Standard Input in the following format:

H W

Output

Print the minimum possible value of S_max−S_min.

Sample Input 1

3 5

Sample Output 1

In the division below, S_max−S_min=5−5=0.

Sample Input 2

4 5

Sample Output 2

In the division below, S_max−S_min=8−6=2.

Sample Input 3

5 5

Sample Output 3

In the division below, S_max−S_min=10−6=4.

Sample Input 4

100000 2

Sample Output 4

Sample Input 5

100000 100000

Sample Output 5

50000


样例倒是十分良心；
我们枚举第一步切的情况；
然后贪心地切剩下部分的中间部分，分为横、竖两种情况；

#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 2000005
#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 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
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 ll rd() {
	ll 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 h, w;
ll minn = 99999999999;
int main() {
//	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> h >> w;
	for (ll i = 1; i < h; i++) {
		ll tmps = i * w;
		ll pos = (h - i) / 2;
		ll tmps1 = (pos)*w;
//		cout << ">>" << tmps1 << endl;
		ll tmps2 = (h - i - pos)*w;
//		cout << ">>"<<tmps2 << endl;
		ll mins = min(min(tmps, tmps1), tmps2);
		ll maxs = max(max(tmps, tmps2), tmps1);
		minn = min(minn, maxs - mins);
		
		pos = w / 2;
		tmps1 = pos * (h - i);
//		cout << ">>" << tmps1 << endl;
		tmps2 = (w - pos)*(h - i);
//		cout << ">>" << tmps2 << endl;
		mins = min(min(tmps, tmps1), tmps2);
		maxs = max(max(tmps, tmps2), tmps1);
		minn = min(minn, maxs - mins);
	//	cout << minn << endl;
	}
	for (ll i = 1; i < w; i++) {
		ll tmps = i * h;
		ll pos = (w - i) / 2;
		ll tmps1 = (pos)*h;
		ll tmps2 = (w - i - pos)*h;
		ll mins = min(min(tmps, tmps1), tmps2);
		ll maxs = max(max(tmps, tmps2), tmps1);
		minn = min(minn, maxs - mins);
	//	cout << minn << endl;
		pos = h / 2;
		tmps1 = (w - i)*pos;
		tmps2 = (w - i)*(h - pos);
		mins = min(min(tmps, tmps1), tmps2);
		maxs = max(max(tmps, tmps2), tmps1);
		minn = min(minn, maxs - mins);
	//	cout << minn << endl;
	}
	cout << minn << endl;
	return 0;
}

转载于:https://www.cnblogs.com/zxyqzy/p/10319198.html