Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4644 [USACO05DEC]Cleaning Shifts S

P4644 [USACO05DEC]Cleaning Shifts S

对于一个贴纸 $[a, b]$ ,有转移:
$$
f _ b = \min _ {k \in [a - 1, b)} f _ k + cost _ i
$$
为了保证转移有序,将所有贴纸按照 $b$ 排序,求区间 $\min$ 可以用线段树优化。

查看代码
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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <climits>
#define inf LLONG_MAX
using namespace std;
typedef long long LL;
const int N = 1e4 + 10, K = 1e5 + 10;
struct Node
{
int l, r;
LL mn;
}tr[K << 2];
struct Cow
{
int l, r;
LL s;
bool operator < (const Cow& x) const
{
return l < x.l;
}
}cow[N];
int n, L, R;
void pushup (int x)
{
tr[x].mn = min(tr[x << 1].mn, tr[x << 1 | 1].mn);
}
void build (int x, int l, int r)
{
tr[x].l = l;
tr[x].r = r;
if (l == r)
{
if (l == L - 1)
tr[x].mn = 0;
else
tr[x].mn = inf;
return;
}
int mid = l + r >> 1;
build(x << 1, l, mid);
build(x << 1 | 1, mid + 1, r);
pushup(x);
}
void modify (int x, int t, LL k)
{
if (tr[x].l == tr[x].r && tr[x].l == t)
{
tr[x].mn = k;
return;
}
int mid = tr[x].l + tr[x].r >> 1;
if (t <= mid)
modify(x << 1, t, k);
else
modify(x << 1 | 1, t, k);
pushup(x);
}
LL query(int x, int l, int r)
{
if (tr[x].l >= l && tr[x].r <= r)
return tr[x].mn;
LL res = inf;
int mid = tr[x].l + tr[x].r >> 1;
if (l <= mid)
res = min(res, query(x << 1, l, r));
if (r > mid)
res = min(res, query(x << 1 | 1, l, r));
return res;
}
int main ()
{
cin >> n >> L >> R;
for (int i = 1; i <= n; i++)
{
cin >> cow[i].l >> cow[i].r >> cow[i].s;
cow[i].l = max(L, cow[i].l);
cow[i].r = min(R, cow[i].r);
}
build(1, L - 1, R);
sort(cow + 1, cow + n + 1);
for (int i = 1; i <= n; i++)
{
LL t = query(1, cow[i].l - 1, cow[i].r - 1);
if (t == inf)
continue;
modify(1, cow[i].r, t + cow[i].s);
}
int t = query(1, R, R);
cout << (t == inf ? -1 : t);
return 0;
}