二维偏序关系 单个值不传递

Posted 2020-10-09 Aragaki

二维偏序关系 只要有一个值比对方大就算

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double EPS = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int  maxn = 1e5 + 100;
const int  maxm = 300;
//next_permutation
//priority_queue<int, vector<int>, greater<int>> que;
int ans[maxn];
int counter[10000005];
struct node
{
        int cf, num, tp;
} person[maxn];
int n, m, i, num[10000001], t[20000001], l, r; //num:原数组；t：树状数组
int lowbit(int x)
{
        return x & (-x); //右起第一个1的位置为第k个 返回值则为2^(k-1)
}
void update(int x, int p) //更新第x 的值
{
        while (x <= 2000000)
        {
                t[x] += p;
                x += lowbit(x); //不断找到自己的祖先 更新值
        }
        return;
}
int sum(int k)//前k个数的和
{
        int ans = 0;
        while (k > 0)
        {
                ans += t[k];
                k -= lowbit(k);
        }
        return ans;
}
int ask(int l, int r) //求l-r区间和
{
        if (r < l)
        {
                return 0;
        }
        return sum(r) - sum(l - 1);
}
bool cmp(node a, node b)
{
        if (a.cf == b.cf)
        {
                return a.tp > b.tp;
        }
        return a.cf < b.cf;
}
void pf()
{
        for (int i = 1; i <= n; i++)
        {
                cout << person[i].cf << " " << person[i].tp << " " << person[i].num << endl;
        }
}
int main()
{
        cin >> n ;
        for (int i = 1; i <= n; i++)
        {
                scanf("%d %d", &person[i].cf, &person[i].tp);
                counter[person[i].cf]++;
                person[i].num = i;
        }
        sort(person + 1, person + 1 + n, cmp);
        //pf();
        //cout << endl;
        for (int i = n; i >= 1; i--)
        {
                int cur = person[i].tp;
                int number = person[i].num;
                update(cur, 1);
                int x = ask(1, cur - 1);
               // cout << "x:" << x << endl;
               // cout << "cf:" << person[i].cf << " " << counter[person[i].cf] << endl;
                ans[number] = i + x - counter[person[i].cf];
                counter[person[i].cf]--;
        }
        for (int i = 1; i <= n; i++)
        {
                cout << ans[i] << endl;
        }
        //    cin>>n>>m;
        //    for(i=1;i<=n;i++)
        //    {
        //        cin>>num[i];
        //        update(i,num[i]);
        //    }
        //    for(i=1;i<=m;i++)
        //    {
        //        cin>>l>>r;
        //        cout<<ask(l,r)<<endl;
        //    }
        return 0;
}