codeforces 629D. Babaei and Birthday Cake

Posted 2020-06-14

题目链接

大意就是给出一个序列， 然后让你从中找出一个严格递增的数列， 使得这一数列里的值加起来最大。

用线段树， 先将数列里的值离散，然后就是线段树单点更新， 区间查询最值。

具体看代码。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
ll v[100005], b[100005], sum[100005*4];
int id[100005];
void pushUp(int rt) {
    sum[rt] = max(sum[rt<<1], sum[rt<<1|1]);
}
void update(int p, ll val, int l, int r, int rt) {
    if(l == r) {
        sum[rt] = val;
        return ;
    }
    int m = l+r>>1;
    if(p<=m)
        update(p, val, lson);
    else
        update(p, val, rson);
    pushUp(rt);
}
ll query(int L, int R, int l, int r, int rt) {
    if(R<L)
        return 0;
    if(L<=l&&R>=r) {
        return sum[rt];
    }
    int m = l+r>>1;
    ll ret = 0;
    if(L<=m)
        ret = query(L, R, lson);
    if(R>m)
        ret = max(ret, query(L, R, rson));
    return ret;
}
int main()
{
    int n, r, h;
    cin>>n;
    for(int i = 0; i<n; i++) {
        scanf("%d%d", &r, &h);
        v[i] = b[i] = 1LL*r*r*h;
    }
    sort(b, b+n);
    int num = unique(b, b+n)-b;
    for(int i = 0; i<n; i++) {
        id[i] = lower_bound(b, b+num, v[i])-b+1;
    }
    ll ans = 0;
    for(int i = 0; i<n; i++) {
        ll tmp = query(1, id[i]-1, 1, num, 1);
        update(id[i], tmp+v[i], 1, num, 1);
        ans = max(ans, tmp+v[i]);
    }
    printf("%.8f", ans*PI);
    return 0;
}