Codeforces Round #343(Div.2) D. Babaei and Birthday Cake(세그먼트 트리 + 이산화 최적화 DP)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const ld eps = 1e-9, PI = 3.1415926535897932384626433832795;
const int mod = 1000000000 + 7;
const int INF = int(1e9);
const ll INF64 = ll(1e18);
const int maxn = 100000 + 10;
int T,n,m;
ll d[maxn],b[maxn];
ll maxv[maxn<<2];
struct node {
    ll r, h, v;
}a[maxn];
void PushUp(int o) {
    maxv[o] = max(maxv[o<<1], maxv[o<<1|1]);
}
void build(int l, int r, int o) {
    int m = (l + r) >> 1;
    maxv[o] = 0;
    if(l == r) return ;
    build(l, m, o<<1);
    build(m+1, r, o<<1|1);
}
void update(int L, int R, int v, int l, int r, int o) {
    int m = (l + r) >> 1;
    if(L <= l && r <= R) {
        if(maxv[o] < d[v]) {
            maxv[o] = d[v];
        }
        return ;
    }
    if(L <= m) update(L, R, v, l, m, o<<1);
    if(m < R) update(L, R, v, m+1, r, o<<1|1);
    PushUp(o);
}
ll query(int L, int R, int l, int r, int o) {
    int m = (l + r) >> 1;
    if(L <= l && r <= R) {
        return maxv[o];
    }
    ll ans = 0;
    if(L <= m) ans = max(ans, query(L, R, l, m, o<<1));
    if(m < R) ans = max(ans, query(L, R, m+1, r, o<<1|1));
    PushUp(o);
    return ans;
}
int main() {
    scanf("%d",&n);
    for(int i=1;i<=n;i++) {
        scanf("%I64d%I64d",&a[i].r,&a[i].h);
        a[i].v = a[i].r * a[i].r * a[i].h;
        b[i] = a[i].v;
    }
    sort(b+1, b+1+n);
    int m = unique(b+1, b+1+n) - b - 1;
    build(1, m, 1);
    ll cur = 0;
    for(int i=1;i<=n;i++) {
        int L = lower_bound(b+1, b+1+m, a[i].v) - b;
        ll v;
        if(L > 1) v = query(1, L-1, 1, m, 1);
        else v = 0;
        d[i] = (v + a[i].v);
        update(L, L, i, 1, m, 1);
        cur = max(cur, d[i]);
    }
    double ans = PI * cur;
    printf("%.8f
",ans);
    return 0;
}