codeforces 540 E. Infinite Inversions (离散化 + 树状数组)

E. Infinite Inversions
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

There is an infinite sequence consisting of all positive integers in the increasing order: p = {1, 2, 3, ...}. We performed n swap operations with this sequence. A swap(a, b) is an operation of swapping the elements of the sequence on positions a and b. Your task is to find the number of inversions in the resulting sequence, i.e. the number of such index pairs (i, j), that i < j and pi > pj.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of swap operations applied to the sequence.

Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 109ai ≠ bi) — the arguments of the swap operation.

Output

Print a single integer — the number of inversions in the resulting sequence.

Sample test(s)
input
2
4 2
1 4
output
4
input
3
1 6
3 4
2 5
output
15

把交换点和交换点之间的区间都当作点,进行离散化,数据量最多达到400000,然后就可以用树状数组处理了

#include 
#include 
#include 
#include 
#include 
using namespace std;

#define P pair
#define PB push_back
#define X first
#define Y second
#define LL __int64

const int N = 1e5 + 5;

vector

pool; int pool_cnt; int u[N], v[N]; int id[N * 4]; int num[N * 4]; int BIT[N * 4]; int lowbit(int x) { return x & -x; } int sum(int x) { int s = 0; while (x > 0) { s += BIT[x]; x -= lowbit(x); } return s; } void update(int x, int c) { while (x <= pool_cnt) { BIT[x] += c; x += lowbit(x); } } int get_p(int x) { return lower_bound(pool.begin(), pool.begin() + pool_cnt, P(x, 0)) - pool.begin() + 1; } int main() { int n; cin>>n; for(int i = 0; i < n; ++i) { cin>>u[i]>>v[i]; pool.PB(P(u[i], 1)); pool.PB(P(v[i], 1)); } sort(pool.begin(), pool.end()); pool_cnt = unique(pool.begin(), pool.end()) - pool.begin(); for(int i = 0; i < pool_cnt - 1; ++i) { int x = pool[i].X + 1; if(x == pool[i + 1].X) continue; int y = pool[i + 1].X - x; pool.PB(P(x, y)); } sort(pool.begin(), pool.end()); pool_cnt = unique(pool.begin(), pool.end()) - pool.begin(); for(int i = 0; i < pool_cnt; ++i) { id[i + 1] = i + 1; num[i + 1] = pool[i].Y; } for(int i = 0; i < n; ++i) { u[i] = get_p(u[i]); v[i] = get_p(v[i]); swap(id[u[i]], id[v[i]]); } LL ans = 0; for(int i = pool_cnt; i > 0; --i) { LL tmp = sum(id[i]); ans += tmp * num[id[i]]; update(id[i], num[id[i]]); } cout<



你可能感兴趣的:(数据结构)