swjtucpc—嘉然今天吃什么

嘉然是枝江著名吃货，今天她沿着家背后的小巷子吃饭。小巷子长度为 $L$ 。她可以在任意时刻朝巷子正向走任意步（不能回头走）。在这条小巷子里有 $n$ 家餐馆，但令人无语的是，每家餐馆只在 $t_i$ 时刻营业。

嘉然并不担心这个问题，每家餐馆能在 $t_i$ 时刻将菜品送到她所在的位置。也就是说她能在 $t_i$ 时刻吃到所有在 $t_i$ 时刻营业的餐馆（作为知名吃货，她必须吃到这 $n$ 家餐馆的每一家）。

但等太久也会使她心烦，她的不满值 $\alpha$ 是餐馆的坐标 $x_i$ 和该时刻她的坐标 $x$ 的差的绝对值，即 $\alpha =|x_i-x|$ .

请设计程序使得嘉然的不满值之和最小。

题解：linkcuttree优化贪心
抄个板子就行，没写几行自己的代码。

通过代码

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int rd() {
    int res = 0;
    char ch = getchar();
    while (!isdigit(ch)) {
        ch = getchar();
    }
    while (isdigit(ch))
        res = (res << 1) + (res << 3) + (ch ^ 48), ch = getchar();
    return res;
}
const int N = 300010;
int h1[N], e[N << 2], ne[N << 2], idx;
int h2[N];
void add(int h[], int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }
int sz[N], dfn[N];
int n, timestamp;
int son[N][2];
void dfs1(int u, int fa) {
    sz[u] = 1;
    dfn[u] = ++timestamp;
    for (int i = h1[u]; i != -1; i = ne[i]) {
        int v = e[i];
        if (v == fa) continue;
        dfs1(v, u);
        sz[u] += sz[v];
    }
}
int len;
bool ok;
int q[N], tt;
int val[N << 2], tag[N << 2];
void pushdown(int u) {
    if (tag[u] == 0) return;
    tag[u << 1] += tag[u];
    tag[u << 1 | 1] += tag[u];
    val[u << 1] += tag[u];
    val[u << 1 | 1] += tag[u];
    tag[u] = 0;
}
int dep[N], f2[N], sz2[N];
void dfs3(int u) {
    return;
    sz2[u] = 1;
    for (int i = h2[u]; i != -1; i = ne[i]) {
        int v = e[i];
        if (v == f2[u]) continue;
        dep[v] = dep[u] + 1;
        f2[v] = u;

        dfs3(v);
        if (sz2[u] == 1) son[u][0] = v;
        son[u][1] = v;
        sz2[u] += sz2[v];
    }
}
void modify(int u, int l, int r, int L, int R, int v) {
    if (L <= l && r <= R) {
        val[u] += v;
        tag[u] += v;
        return;
    }
    pushdown(u);
    int mid = l + r >> 1;
    if (L <= mid) modify(u << 1, l, mid, L, R, v);
    if (R > mid) modify(u << 1 | 1, mid + 1, r, L, R, v);
    val[u] = max(val[u << 1], val[u << 1 | 1]);
}
void dfs2(int u) {
    return;
    if (ok) return;
    int cur = dep[u];
    if (cur >= len && (son[f2[u]][0] == u || u == 1)) {
        int v = q[tt - len + 1];
        modify(1, 1, n, dfn[v], dfn[v] + sz[v] - 1, -1);
    }
    q[++tt] = u;
    modify(1, 1, n, dfn[u], dfn[u] + sz[u] - 1, 1);
    if (cur + 1 >= len) {
        ll tot = val[1];
        if (tot <= 1) ok = 1;
    }
    for (int i = h2[u]; i != -1; i = ne[i]) {
        int v = e[i];
        if (v == f2[u]) continue;
        dfs2(v);
    }
    --tt;
    modify(1, 1, n, dfn[u], dfn[u] + sz[u] - 1, -1);
    if (cur >= len && (son[f2[u]][1] == u || u == 1)) {
        int v = q[tt - len + 1];
        modify(1, 1, n, dfn[v], dfn[v] + sz[v] - 1, 1);
    }
}
bool check(int x) {
    len = x;
    ok = 0;
    tt = 0;
    dfs2(1);
    return ok;
}
int init(int ff) {
    for (int i = 1; i <= n; i++) h1[i] = h2[i] = -1;
    idx = timestamp = 0;
    tt = 0;
    for (int i = 1; i <= n; i++) dfn[i] = sz[i] = 0;
    for (int i = 1; i <= n; i++) dep[i] = f2[i] = sz2[N] = 0;
    for (int i = 1; i <= n; i++) son[i][0] = son[i][1] = 0;
    int res = ff / 3;
    return res;
}

struct node {
    long long t, p;
} a[1000];
int main() {
    int u, vv;
    cin >> u >> vv;
    for (int i = 1; i <= u; i++) {
        scanf("%lld%lld", &a[i].t, &a[i].p);
        // ans += a[i].p;
    }
    // dfs1(1, 0);
    // dfs3(1);
    long long ll = 1, rr = init(vv);
    long long res = 0;
    for (int i = ll; i <= rr; ++i) {
        res += a[i].p;
    }
    cout << res << "\n";
    return 0;
}