【USACO3.1.2】总分 背包问题动态规划

完全背包问题。

f[i][j]表示用前j个物品，放满i的空间，能取得最大价值。 f[i][j] =max  f[i - a[P]]j - 1] + b[P]  

j按顺序用的话，数组第二维可以压掉。即为f[i] = max f[i - a[P]] + b[P]

Executing...
   Test 1: TEST OK [0.008 secs, 3484 KB]
   Test 2: TEST OK [0.008 secs, 3484 KB]
   Test 3: TEST OK [0.008 secs, 3484 KB]
   Test 4: TEST OK [0.005 secs, 3484 KB]
   Test 5: TEST OK [0.019 secs, 3484 KB]
   Test 6: TEST OK [0.032 secs, 3484 KB]
   Test 7: TEST OK [0.057 secs, 3484 KB]
   Test 8: TEST OK [0.119 secs, 3484 KB]
   Test 9: TEST OK [0.230 secs, 3484 KB]
   Test 10: TEST OK [0.221 secs, 3484 KB]
   Test 11: TEST OK [0.003 secs, 3484 KB]
   Test 12: TEST OK [0.003 secs, 3484 KB]

All tests OK.

n^2的算法居然挺快的。

大量读入，getcahr目测要快更多？

/*
TASK:inflate
LANG:C++
*/
#include <cstdio>
#include <cstring>
#define max(a,b) ((a)>(b)?(a):(b))
int m, n;
int a[10001], b[10001];
int c[10001], ans = -1;
int main()
{
	freopen("inflate.in","r",stdin);
	freopen("inflate.out","w",stdout);
	scanf("%d%d", &m, &n);
	for (int i = 0; i != n; ++ i)	scanf("%d%d", &a[i], &b[i]);//价值 体积
	memset(c, -1, sizeof(c));
	c[0] = 0;
	for (int i = 0; i != n; ++ i)
		for (int j = 0; j <= m-b[i]; ++ j)
		{
			if (c[j]!=-1)	
			{
				c[j+b[i]] = max(c[j+b[i]], c[j] + a[i]);
				ans = max(c[j + b[i]], ans);
			}
		}
	printf("%d\n", ans);
	return 0;
}