CF797F：Mice and Holes（dp + 单调队列优化）

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本文探讨了一道算法题目，即如何计算一群老鼠进入一系列有容量限制的洞中时，所有老鼠进入洞穴路径总长度的最小值。文章提供了一个使用动态规划与单调队列优化的解决方案。

One day Masha came home and noticed n mice in the corridor of her flat. Of course, she shouted loudly, so scared mice started to run to the holes in the corridor.

The corridor can be represeted as a numeric axis with n mice and m holes on it. ith mouse is at the coordinate xi, and jth hole — at coordinate pj. jth hole has enough room for cj mice, so not more than cj mice can enter this hole.

What is the minimum sum of distances that mice have to go through so that they all can hide in the holes? If ith mouse goes to the hole j, then its distance is |xi - pj|.

Print the minimum sum of distances.

Input

The first line contains two integer numbers n, m (1 ≤ n, m ≤ 5000) — the number of mice and the number of holes, respectively.

The second line contains n integers x1, x2, ..., xn ( - 109 ≤ xi ≤ 109), where xi is the coordinate of ith mouse.

Next m lines contain pairs of integer numbers pj, cj ( - 109 ≤ pj ≤ 109, 1 ≤ cj ≤ 5000), where pj is the coordinate of jth hole, and cj is the maximum number of mice that can hide in the hole j.

Output

Print one integer number — the minimum sum of distances. If there is no solution, print -1 instead.

Examples

input
4 5
6 2 8 9
3 6
2 1
3 6
4 7
4 7

output
11

input
7 2
10 20 30 40 50 45 35
-1000000000 10
1000000000 1

output
7000000130

题意：X坐标轴上，有N只老鼠，M个鼠洞，每个鼠洞有一定的容量，求所有老鼠进入鼠洞路径和的最小值。

思路：首先需要知道，最优情况下老鼠回洞的路径是不会发生交叉的，那么可以用dp解决，题解是这样的：dp[i][j]表示前i个洞被j只老鼠占据的最小步数。dp[i][j] = min(dp[i-1][k]+sum[i][j]-sum[i][k])，枚举k从0到j，那么O(n^3)是超时的，考虑到sum[i][j]是定值，且sum[i][k]多次重复枚举了，可以用单调队列优化下。---第一次用List超时了，看来还是用数组模拟快一些。

//reference : meopass
# include <iostream>
# include <cstring>
# include <cstdlib>
# include <algorithm>
# include <cstdio>
using namespace std;
typedef long long LL;
struct node
{
    LL x, v;
    bool operator < (const node&a) const
    {
        return x < a.x;
    }
}hole[5003];
LL mice[5003], sum[5003], pre[5003], dp[5003][5003];
int que[5003];
const int INF = 0x3f3f3f3f;
int main()
{
    int n, m;
    memset(dp, INF, sizeof(dp));
    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; ++i)
        scanf("%I64d",&mice[i]);
    for(int i=1; i<=m; ++i)
        scanf("%I64d%I64d",&hole[i].x, &hole[i].v);
    sort(mice+1, mice+n+1);
    sort(hole+1, hole+m+1);
    for(int i=1; i<=m; ++i)
        pre[i] = pre[i-1] + hole[i].v;//洞容纳量前缀和，装不下所有老鼠直接输出-1。
    if(pre[m] < n) return 0*puts("-1");
    dp[0][0] = 0;
    for(int i=1; i<=m; ++i)
    {
        dp[i][0] = 0;
        int l=0, r=0;
        que[++r] = 0;
        for(int j=1; j<=pre[i]&& j<=n; ++j)
        {
            dp[i][j] = dp[i-1][j];
            sum[j] = sum[j-1] + llabs(mice[j]-hole[i].x);
            while(j-que[l]>hole[i].v) ++l;
            while(l<=r && dp[i-1][que[l]]-sum[que[l]] > dp[i-1][j]-sum[j]) ++l;;
            que[++r] = j;
            dp[i][j] = min(dp[i][j], (LL)sum[j]+dp[i-1][que[l]]-sum[que[l]]);
        }
    }
    printf("%I64d\n",dp[m][n]);
}