[luoguP2224] [HNOI2001]产品加工（背包DP）

Posted 2020-10-21 蒟蒻zht的博客

传送门

 

f[i][j]表示第一个机器耗时j，第二个机器耗时f[i][j]

第一维可以滚掉

#include <cstdio>
#include <cstring>
#include <iostream>
#define INF 1e9
#define min(x, y) ((x) < (y) ? (x) : (y))
#define max(x, y) ((x) > (y) ? (x) : (y))

using namespace std;

int n, m, now = 0, last = 1, ans = 1e9;
int f[2][30011];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
	return x * f;
}

int main()
{
	int i, j, x, y, z;
	n = read();
	for(i = 1; i <= n; i++)
	{
		x = read(), y = read(), z = read();
		m += max(x, max(y, z));
		if(!x) x = INF;
		if(!y) y = INF;
		if(!z) z = INF;
		now ^= 1, last ^= 1;
		memset(f[now], 127 / 3, sizeof(f[now]));
		for(j = m; j >= 0; j--)
		{
			if(j >= x) f[now][j] = min(f[now][j], f[last][j - x]);
			f[now][j] = min(f[now][j], f[last][j] + y);
			if(j >= z) f[now][j] = min(f[now][j], f[last][j - z] + z);
		}
	}
	for(i = m; i >= 0; i--)
		ans = min(ans, max(i, f[now][i]));
	printf("%d\n", ans);
	return 0;
}