CodeForces - 38E Let's Go Rolling!

Description

On a number axis directed from the left rightwards, n marbles with coordinates
x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that
is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number
i is equal to ci, number
ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't
move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you
will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:

the sum of the costs of stuck pins;
the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.

Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.

Input

The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next
n lines contain the descri_ptions of the marbles in pairs of integers
xi,
ci ( - 109 ≤ xi, ci ≤ 109).
The numbers are space-separated. Each descri_ption is given on a separate line. No two marbles have identical initial positions.

Output

Output the single number — the least fine you will have to pay.

Sample Input

Input

3
2 3
3 4
1 2

Output

5

Input

4
1 7
3 1
5 10
6 1

Output

11

题意：给你每个求的位置以及在这个点修卡点的费用，每个球都会向左滚，求让所有球停下来的最小费用

思路：DP， 先处理出当前点之前都跑到第一个点的距离和，然后用dp[i]表示到第i个点的最小费用，假设停在了第j个点，那么我们就要算上[j, i]所有点都跑到j的距离和，还有要算上修j的费用，最后减去[j, i]要跑到第1点的距离

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long ll;
const ll inf = 1e15;
const int maxn = 3005;

struct Node {
	int x, c;
	bool operator <(const Node &a) const {
		return x < a.x;
	}
} node[maxn];

ll sum[maxn], dp[maxn];

int main() {
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d%d", &node[i].x, &node[i].c);
	sort(node+1, node+1+n);
	sum[0] = 0;
	memset(dp, 0, sizeof(dp));
	for (int i = 1; i <= n; i++)	
		sum[i] = sum[i-1] + node[i].x - node[1].x;
	for (int i = 1; i <= n; i++) {
		dp[i] = inf;
		for (int j = 1; j <= i; j++)
			dp[i] = min(dp[i], dp[j-1]+sum[i]-sum[j-1]-(ll)(node[j].x-node[1].x)*(i-j+1)+node[j].c);
	}	
	printf("%lld\n", dp
);
	return 0;
}