D - Nested Segments
You are given n segments on a line. There are no ends of some segments that coincide. For each segment find the number of segments it contains.
Input
The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of segments on a line.
Each of the next n lines contains two integers l**i and r**i ( - 109 ≤ l**i < r**i ≤ 109) — the coordinates of the left and the right ends of the i-th segment. It is guaranteed that there are no ends of some segments that coincide.
Output
Print n lines. The j-th of them should contain the only integer a**j — the number of segments contained in the j-th segment.
Examples
Input
41 82 34 75 6
Output
3010
Input
33 41 52 6
Output
011
题意：
给你n个线段，
保证没有两个线段的端点是相同的。
问你每一个线段包含了多少个其他的线段？
思路：
离散化+树状数组
首先对给定的坐标进行离散化，那么所有坐标都在 2*n的范围内了。
这个范围是树桩数组可以维护的。
然后我们按照x坐标降序排序。
然后从1~n依次扫描线段，
对于每一个线段，用树状数组询问1~y 的sum和就是它包含的线段个数。
然后把自己的y坐标在树桩数组中+1
因为询问时已经在树状数组中的线段的x都比当前线段的x大，所以只需要查看有多少个y坐标小于当前的y坐标即可。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
// #define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
struct pii {
int fi;
int se;
int id;
};
pii a[maxn];
std::vector<int> v;
int n;
int tree[maxn];
int lowbit(int x)
{
return -x & x;
}
void add(int x, int val)
{
while (x < maxn) {
tree[x] += val;
x += lowbit(x);
}
}
int ask(int x)
{
int res = 0;
while (x) {
res += tree[x];
x -= lowbit(x);
}
return res;
}
pii b[maxn];
int ans[maxn];
bool cmp(pii aa, pii bb)
{
if (aa.fi != bb.fi) {
return aa.fi > bb.fi;
} else {
return aa.se < bb.se;
}
}
int main()
{
//freopen("D:\code\text\input.txt","r",stdin);
//freopen("D:\code\text\output.txt","w",stdout);
cin >> n;
repd(i, 1, n) {
cin >> a[i].fi >> a[i].se;
a[i].id = i;
v.push_back(a[i].fi);
v.push_back(a[i].se);
}
sort(ALL(v));
repd(i, 1, n) {
b[i].fi = lower_bound(ALL(v), a[i].fi) - v.begin() + 1;
b[i].se = lower_bound(ALL(v), a[i].se) - v.begin() + 1;
b[i].id = a[i].id;
}
sort(b + 1, b + 1 + n, cmp);
repd(i, 1, n) {
ans[b[i].id] = ask(b[i].se);
add(b[i].se, 1);
}
repd(i, 1, n) {
cout << ans[i] << endl;
}
return 0;
}
inline void getInt(int *p)
{
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '
');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
} else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}