CodeForces - 38E Let's Go Rolling!

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Description

On a number axis directed from the left rightwards, n marbles with coordinatesx_{1,?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 numberi is equal to ci, numberci 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 nextn lines contain the descri_ptions of the marbles in pairs of integersx_{i,ci (?-?10^{9?≤?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

32 33 41 2

Output

5

Input

41 73 15 106 1

Output

11

題意：給你每個求的位置以及在這個點修卡點的費用，每個球都會向左滾，求讓所有球停下來的最小費用

思路：DP， 先處理出當前點之前都跑到第一個點的距離和，然後用dp[i]表示到第i個點的最小費用，假設停在了第j個點，那麼我們就要算上[j, i]所有點都跑到j的距離和，還有要算上修j的費用，最後減去[j, i]要跑到第1點的距離

#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[n]);return 0;}