F. Greetings

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-062.C. Removal of Unattractive Pairs2023-12-063.D. Jumping Through Segments2023-12-084.E. Good Triples2023-12-085.F. Shift and Reverse2023-12-086.D. Yet Another Monster Fight2023-12-117.A. Constructive Problems2023-12-188.C. Game with Multiset2023-12-199.A. Rating Increase2023-12-1910.B. Swap and Delete2023-12-1911.A. Problemsolving Log2023-12-2012.B. Preparing for the Contest2023-12-2013.C. Quests2023-12-2014.D. Three Activities2023-12-2015.E2. Game with Marbles (Hard Version)2023-12-2016.cf刷题有感2023-12-2017.A. Anonymous Informant2023-12-2118.A. Forked!2023-12-2319.B. Make Almost Equal With Mod2023-12-2720.C. Heavy Intervals2023-12-2721.D. Split Plus K2023-12-2822.A. 20232023-12-3123.B. Two Divisors2023-12-3124.C. Training Before the Olympiad2023-12-3125.D. Mathematical Problem2024-01-02

26.F. Greetings2024-01-02

27.C. Partitioning the Array2024-01-1528.G. Bicycles2024-01-1629.E. Eat the Chip2024-01-1730.G. Lights2024-01-1731.D. Array Repetition2024-01-1732.D. Berserk Monsters2024-01-2233.E. Increasing Subsequences2024-01-2334.D. Very Different Array2024-01-2435.G. Mischievous Shooter2024-01-2536.B. Plus-Minus Split2024-01-2737.B. A Balanced Problemset2024-01-2838.C. Did We Get Everything Covered2024-01-2839.D. Find the Different Ones!2024-02-0840.C. Grouping Increases2024-02-1041.D. Good Trip2024-02-1242.C. Physical Education Lesson2024-02-1243.E. Final Countdown2024-02-1844.D. Divisible Pairs2024-02-1945.G. Vlad and Trouble at MIT2024-02-2046.A. Brick Wall2024-02-2847.B. Minimize Inversions2024-02-2848.C. XOR-distance2024-02-2849.A. Moving Chips2024-02-2850.B. Monsters Attack!2024-02-2851.C. Find B2024-02-2852.D. Slimes2024-02-2853.C. Turtle Fingers: Count the Values of k2024-02-2854.D. Turtle Tenacity: Continual Mods2024-02-2855.D. Vlad and Division2024-02-2856.C. LR-remainders2024-02-2857.D. Lonely Mountain Dungeons2024-03-12

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原题链接

读题

所有人的速度相同，也就是说，如果大家都在动，那么大家永不相交。所以相交的情况有且仅有b到终点了，而a还在b的后面，且a的终点在b的后面

将上述情况模型化后，就是求每个人行动的区间包含了多少人，然后对每个人求和

题解

再度简化，对于第$i$人来说，就是求$r$小于它且$l$大于它的人数

我们可以这样想，对r进行升序排序，然后遍历，这个时候$i-1$所代表的是$r$小于其的人数，然后怎么求r小于它的人里l大于它的呢？
天才的想法：反过来求，利用前缀和，找出l小于它的人数，然后相减就是答案。前缀和用树状数组快速求解

代码

#include<bits/stdc++.h>
#define ll long long
using namespace std;
struct node
{
    ll l;
    ll r;
    friend bool operator <(node a,node b){return a.r<b.r;}
}edge[200005];
ll l[200005]={0};
ll pres[200005]={0};
ll n;
ll lowbit(ll x)
{
    return x & -x;
}
void occur(ll x)
{
    for(ll i=x;i<=n;i+=lowbit(i))pres[i]++;
}
ll inquire(ll x)
{
    ll cnt=0;
    for(ll i=x;i>=1;i-=lowbit(i))cnt+=pres[i];
    return cnt;
}
int main()
{
    ios_base::sync_with_stdio(false);
    ll t;
    cin>>t;
    while(t--)
    {
        memset(pres,0,sizeof(pres));
        cin>>n;
        for(ll i=1;i<=n;i++)
        {
            cin>>l[i];
            cin>>edge[i].r;
            edge[i].l=l[i];
        }
        sort(l+1,l+n+1);
        sort(edge+1,edge+1+n);
        ll ans=0;
        for(ll i=1;i<=n;i++)
        {
            ll pos=lower_bound(l+1,l+1+n,edge[i].l)-l;
            ll pre=inquire(pos);
            ans+=i-1-pre;
            occur(pos);
        }
        cout<<ans<<endl;
    }
    return 0;
}