(中等) CF 311B Cats Transport，斜率优化DP。

　　Zxr960115 is owner of a large farm. He feeds m cute cats and employs p feeders. There's a straight road across the farm and n hills along the road, numbered from 1 to n from left to right. The distance between hill i and (i - 1) is d_i meters. The feeders live in hill 1.

One day, the cats went out to play. Cat i went on a trip to hill h_i, finished its trip at time t_i, and then waited at hill h_i for a feeder. The feeders must take all the cats. Each feeder goes straightly from hill 1 to n without waiting at a hill and takes all the waiting cats at each hill away. Feeders walk at a speed of 1 meter per unit time and are strong enough to take as many cats as they want.

For example, suppose we have two hills (d₂ = 1) and one cat that finished its trip at time 3 at hill 2 (h₁ = 2). Then if the feeder leaves hill 1 at time 2 or at time 3, he can take this cat, but if he leaves hill 1 at time 1 he can't take it. If the feeder leaves hill 1 at time 2, the cat waits him for 0 time units, if the feeder leaves hill 1 at time 3, the cat waits him for 1 time units.

Your task is to schedule the time leaving from hill 1 for each feeder so that the sum of the waiting time of all cats is minimized.

 

　　队伍训练时做到的题目，好不容易推出了dp公式并且想到了斜率优化，结果犯了SB错误导致一直错。。。

　　题目大致就是几只猫在一些地方，然后让人去收猫。。。饿还是看题吧。。。

　　直接对于某个位置的猫的 t 减去到这个位置的时间就好，问题就转换成了有一些猫每个猫都有一个值，然后给P个人分别一个值，然后每个猫的找到比他大的最近的那个人的值，然后相减，累加每个猫的，让总和最小。。。饿，表达稍微比较烂。。。

　　这里考虑到每个人的值一定是某只猫的值H，不然向下移动一点点可以更优。

　　对每个猫的值进行排序，然后从左到右dp，

　　dp[i][j]表示前i个猫，使用了j个人，的最小值。。。

　　然后递推的话 dp[i][j]=min{ dp[x][j-1]+(i-x)H[i]-S[i]+S[x] };

　　S表示H的前缀和。

　　然后转换一下变成 min{ dp[x][j-1]+S[x]-xH[i] } + iH[i]-S[i];

　　然后 Y[x]=dp[x][j-1] ,X[x]=x;

　　然后就是经典斜率DP的问题了。。。

代码如下：

// ━━━━━━神兽出没━━━━━━
//      ┏┓       ┏┓
//     ┏┛┻━━━━━━━┛┻┓
//     ┃           ┃
//     ┃     ━     ┃
//     ████━████   ┃
//     ┃           ┃
//     ┃    ┻      ┃
//     ┃           ┃
//     ┗━┓       ┏━┛
//       ┃       ┃
//       ┃       ┃
//       ┃       ┗━━━┓
//       ┃           ┣┓
//       ┃           ┏┛
//       ┗┓┓┏━━━━━┳┓┏┛
//        ┃┫┫     ┃┫┫
//        ┗┻┛     ┗┻┛
//
// ━━━━━━感觉萌萌哒━━━━━━

// Author        : WhyWhy
// Created Time  : 2015年10月09日 星期五 18时45分52秒
// File Name     : B.cpp

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;

const int MaxN=100005;

int N,M,P;
long long DP1[MaxN],DP2[MaxN];
long long *dp1,*dp2;
long long H[MaxN];
long long S[MaxN];

long long X[MaxN],Y[MaxN],cou;

long long d[MaxN];

bool better(int a,int b,long long H)
{
    return (Y[a]-X[a]*H)<=(Y[b]-X[b]*H);
}

bool judge(long long x1,long long y1,long long x2,long long y2,long long x3,long long y3)
{
    return (y1-y2)*(x2-x3)<=(y2-y3)*(x1-x2);
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    ios::sync_with_stdio(false);
    cin>>N>>M>>P;

    d[1]=0;
    for(int i=2;i<=N;++i)
    {
        cin>>d[i];
        d[i]+=d[i-1];
    }
    long long a,b;
    for(int i=1;i<=M;++i)
    {
        cin>>a>>b;
        H[i]=b-d[a];
    }

    dp1=DP1;
    dp2=DP2;
    N=M;
    sort(H+1,H+N+1);

    S[0]=0;
    for(int i=1;i<=N;++i)
    {
        S[i]=S[i-1]+H[i];
        dp1[i]=H[i]*i-S[i];
    }

    int p;
    long long TX,TY;
    P=min(P,N);

    for(int j=2;j<=P;++j)
    {
        cou=1;
        Y[0]=dp1[j-1]+S[j-1];
        X[0]=j-1;
        p=0;

        for(int i=j;i<=N;++i)
        {
            while(p<cou-1 && better(p+1,p,H[i])) ++p;
            dp2[i]=Y[p]-X[p]*H[i]-S[i]+i*H[i];

            TX=i;
            TY=dp1[i]+S[i];

            while(cou-1>p && judge(TX,TY,X[cou-1],Y[cou-1],X[cou-2],Y[cou-2])) --cou;
            X[cou]=TX;
            Y[cou++]=TY;
        }
        swap(dp1,dp2);
    }

    cout<<dp1[N]<<endl;

    return 0;
}