POJ2184 Cow Exhibition 背包

题目大意：已知c[i]...c[n]及f[i]...f[n]，现要选出一些i，使得当sum{c[i]}和sum{f[i]}均非负时，sum(c[i]+f[i])的最大值。

以sum(c[i])（c[i]>=0）作为背包容量totV，i当作物体，c[i]当作物体体积，f[i]当作物体重量，01背包后,求max{j+DP[j]} （非负）即可。

注意：

• 这里物体体积存在负数，所以循环j时起始条件不能为j=0！而应当在for下面加if，判断子问题在minJ~maxJ范围之内。
• 初值不是DP[最左端点]=0，而是DP[0点所在位置]=0。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_V = 200010, MAX_OBJ = 110;
#define M(x) x+offset//move

int DP(int minJ, int maxJ, int totObj, int *_v, int *_w)
{
	static int DP[MAX_V];
	int offset = max(-minJ, 0);
	memset(DP, 0xcf, sizeof(DP));
	DP[M(0)] = 0;
	for (int i = 1; i <= totObj; i++)
	{
		if (_v[i] <= 0)
		{
			for (int j = minJ; j <= maxJ; j++)
				if (j >= minJ + _v[i] && j <= maxJ + _v[i])
					DP[M(j)] = max(DP[M(j)], DP[M(j - _v[i])] + _w[i]);
		}
		else
		{
			for (int j = maxJ; j >= minJ; j--)
				if (j >= minJ + _v[i] && j <= maxJ + _v[i])
					DP[M(j)] = max(DP[M(j)], DP[M(j - _v[i])] + _w[i]);
		}
	}
	int ans = 0;
	for (int j = 0; j <= maxJ; j++)
		if(DP[M(j)] >=0)
		  ans = max(ans, j+DP[M(j)]);
	return ans;
}

int main()
{
#ifdef _DEBUG
	freopen("c:\\noi\\source\\input.txt", "r", stdin);
#endif
	static int _s[MAX_OBJ], _f[MAX_OBJ];
	int n, minS = 0, maxS = 0, cnt = 0, s, f, tempAns = 0;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d%d", &s, &f);
		if (s >= 0 && f >= 0)
		{
			tempAns += s + f;
			continue;
		}
		else if (s <= 0 && f <= 0)
			continue;
		else
		{
			cnt++;
			_s[cnt] = s;
			_f[cnt] = f;
			_s[cnt] > 0 ? maxS += _s[cnt] : minS += _s[cnt];

		}
	}
	printf("%d\n", tempAns + DP(minS, maxS, cnt, _s, _f));
	return 0;
}