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题目大意

有一些猫，放在一些位置，人一步移动一个位置
给出每个猫出现的时间，每个人可以自由安排其出发时间，沿途已经出现的猫捡起，猫等待的时间是被减去的时间减去出现的时间
猫可以等人，人不能等猫
现在有P个人，问猫等待的总时间T最小是多少。

思路

首先可以算出要捡起每个猫最早出发的时间是多少\(c_i\)
然后按照这个值排序，每个人捡起的猫一定是在这个值上连续的一段
然后计算出如果一个人选\([l,r]\)，等待时间到达每个猫的时间减去到个猫最小的时间，是\(c_r*(r-l+1)-(s_r-s_{l-1})\)
其中s是c的前缀和，为什么呢？因为c是刚好捡起没个猫的时间，所以\(c_r-c_i\)就是刚好捡起r时候i的等待时间
然后就变成了\(dp_{i,j}=dp_{k,j-1}+c_i*(i-k)-(s_i-s_k)\)
转化一下发现\(x'=k,k'=c_i,y'=dp_{k,j-1}+s_k,b'=dp_{i,j}+s_i-i*c_i\)
斜率单调递增，最小化截距，直接维护下凸壳就可以了

//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
//typename
typedef long long ll;
typedef pair<int, int> pi;
//convenient for
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
//inf of different typename
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
//fast read and write
template <typename T>
void Read(T &x) {bool w = 1;x = 0;char c = getchar();while (!isdigit(c) && c != '-') c = getchar();if (c == '-') w = 0, c = getchar();while (isdigit(c)) {x = (x<<1) + (x<<3) + c -'0';c = getchar();}if (!w) x = -x;
}
template <typename T>
void Write(T x) {if (x < 0) {putchar('-');x = -x; }if (x > 9) Write(x / 10);putchar(x % 10 + '0');
}
//----------------------------------------------
//c_i = t_i - d_i
//s_i = \sum_{j=1}^i c_j
//dp[i][j] = min(dp[k][j - 1] + c_i * (i - k) - (s_i - s_k)
const int N = 1e5 + 10;
const int K = 1e2 + 10;
ll n, m, p;
ll d[N], s[N], c[N];
ll dp[N][K];
struct Node {ll x, y;} q[N];
ll operator * (const Node a, const Node b) {return a.x * b.y - a.y * b.x;
}
Node operator - (const Node a, const Node b) {return (Node){a.x - b.x, a.y - b.y};
}
ll calc(Node a, ll pos) {return a.y - s[pos] + c[pos] * (pos - a.x);
}
int main() {freopen("input.txt", "r", stdin);Read(n), Read(m), Read(p);fu(i, 2, n) Read(d[i]), d[i] += d[i - 1];fu(i, 1, m) {int t, h; Read(h), Read(t);c[i] = t - d[h];}sort(c + 1, c + m + 1);fu(i, 1, m) {s[i] = s[i - 1] + c[i];dp[i][1] = - s[i] + i * c[i];}int l, r;fu(j, 2, p) {q[l = r = 1] = (Node) {0ll, 0ll};fu(i, 1, m) {while (l < r && calc(q[l], i) >= calc(q[l + 1], i)) ++l;dp[i][j] = calc(q[l], i);Node now = (Node){i, dp[i][j - 1] + s[i]};while (l < r && (q[r] - now) * (q[r - 1] - now) > 0) --r;q[++r] = now;}}Write(dp[m][p]);return 0;
}