SPOJ 345 Mixtures (区间DP&前缀数组)

http://www.spoj.com/problems/MIXTURES/

简单区间DP，注意用前缀数组来计算从i到j的连续物品的混合颜色值。

dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + ((sum[k] - sum[i - 1]) % 100) * ((sum[j] - sum[k]) % 100));

完整代码：

/*0.03s, 2.7M*/

#include<bits/stdc++.h>
using namespace std;

int dp[101][101], sum[101];

int main()
{
	int n, i, j, k, len;
	while (~scanf("%d", &n))
	{
		memset(dp, 0, sizeof(dp));
		memset(sum, 0, sizeof(sum));
		for (i = 1; i <= n; ++i)
			scanf("%d", &sum[i]), sum[i] += sum[i - 1];///前缀数组，这样sum[j]-sum[i-1]就相当于从i到j的连续物品的混合颜色值
		for (len = 1; len < n; ++len)
			for (i = 1; i <= n - len; ++i)
			{
				j = i + len;
				dp[i][j] = INT_MAX;
				for (k = i; k < j; ++k)
					dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + ((sum[k] - sum[i - 1]) % 100) * ((sum[j] - sum[k]) % 100));
			}
		printf("%d\n", dp[1][n]);
	}
	return 0;
}