HDU 1300

Dp[i]=min{Dp[j]+V}, 0<=j<i, V为第j+1类珠宝到第i类全部以i类买入的价值;

题目默认每个输入的价格递增，如果非递增，会麻烦一些。

#include <iostream>
#include <string.h>
using namespace std;
const int N = 105;
const int INF = 9999999;
int dp[N];
int a[N];
int p[N];

int sum(int x, int y, int p)
{
	int num = 0; 
	for(int i = x; i <= y; i++)
		num += a[i];
	return (num + 10) * p;
}
int main()
{
	int t;
	cin >> t;
	while(t--)
	{
		memset(dp, INF, sizeof(dp));
		memset(a, 0, sizeof(a));
		int n;
		cin >> n;
		for(int i = 1; i <= n; i++)
		{
			cin >> a[i] >> p[i];
		}
		dp[0] = 0;
		//dp[1] = (a[1] + 10) * p[1];
		for(int i = 1; i <= n; i++)
		{
			for(int j = 0; j < i; j++)
			{
				dp[i] = min(dp[i], dp[j] + sum(j+1, i, p[i]));
			}
		}
		cout << dp[n]  << endl;
	}
	return 0;
}