RT。

这道题我的大致思路是设 $f_i$ 表示完全覆盖 $[L,i]$ 时的最小花费，首先将所有工作时间按照右端点 $r$ 升序排序，那么对于第 $i$ 头牛有转移方程：$f_{a_i.r}=\min\limits_{a_i.l-1\leq j \leq a_i.r-1}(f_j) + a_i.val$，$val$ 是花费，使用线段树对取最小值这一块优化。

根据目前初步调试，疑似是线段树的单点修改 change 和区间查询 ask 出了问题（当然不排除其他地方出错的可能性）。

调了很久了 /kk，有谁能帮帮我吗 /kel

/*
========= Plozia =========
    Author:Plozia
    Problem:P4644 [USACO05DEC]Cleaning Shifts S
    Date:2021/5/6
========= Plozia =========
*/

#include <bits/stdc++.h>

typedef long long LL;
const int MAXN = 10000 + 10, MAXT = 86399 + 10;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int n, s, t;
LL f[MAXT];
struct node { int l, r; LL val; } a[MAXN];
struct node1
{
    int l, r;
    LL minn;
    #define l(p) tree[p].l
    #define r(p) tree[p].r
    #define m(p) tree[p].minn
}tree[MAXT << 2];

int read()
{
    int sum = 0, fh = 1; char ch = getchar();
    for (; ch < '0' || ch > '9'; ch = getchar()) fh -= (ch == '-') << 1;
    for (; ch >= '0' && ch <= '9'; ch = getchar()) sum = (sum << 3) + (sum << 1) + (ch ^ 48);
    return sum * fh;
}
bool cmp(const node &fir, const node &sec) { return fir.r < sec.r; }
LL Min(LL fir, LL sec) { return (fir < sec) ? fir : sec; }
LL Max(LL fir, LL sec) { return (fir > sec) ? fir : sec; }

void build(int p, int l, int r)//建树
{
    l(p) = l, r(p) = r;
    if (l == r) { m(p) = INF; return ; }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid); build(p << 1 | 1, mid + 1, r);
}

void change(int p, int x, LL d)//单点修改
{
    if (l(p) == r(p) && l(p) == x) { m(p) = Min(m(p), d); return ;}
    int mid = (l(p) + r(p)) >> 1;
    if (x <= mid) change(p << 1, x, d);
    else change(p << 1 | 1, x, d);
    m(p) = Min(m(p << 1), m(p << 1 | 1));
}

LL ask(int p, int l, int r)//区间查询
{
    if (l(p) >= l && r(p) <= r) return m(p);
    int mid = (l(p) + r(p)) >> 1; LL val = INF;
    if (l <= mid) val = Min(val, ask(p << 1, l, r));
    if (r > mid) val = Min(val, ask(p << 1 | 1, l, r));
    return val;
}

int main()
{
    n = read(), s = read() + 1, t = read() + 1;
    for (int i = 1; i <= n; ++i) a[i].l = Max(read() + 1, s), a[i].r = Min(read() + 1, t), a[i].val = (LL)read();
    std::sort(a + 1, a + n + 1, cmp);//排序
    build(1, 0, t); memset(f, 0x3f, sizeof(f)); f[s - 1] = 0; change(1, s - 1, 0);//初始化
    for (int i = 1; i <= n; ++i)
    {
        LL sum = ask(1, a[i].l - 1, a[i].r - 1);
        if (sum == INF) continue ;
        f[a[i].r] = sum + a[i].val;
        change(1, a[i].r, f[a[i].r]);
    }//转移
    if (f[t] == INF) printf("-1\n");
    else printf("%lld\n", f[t]);
    return 0;
}