CF 540E, 树状数组

题目大意：在1~10^9的范围内随便交换某些位置上的数，求逆序对数量，交换位置<=10^5

解：因为是交换位置很少，离散化来做，逆序对可以看成两部分，一部分是出现位置的逆序对，另一部分的出现了的数对于没有交换位置上的数（没有在离散化中出现的数）的逆序对。分别统计一下，第一part用树状数组，第二part之间算一下区间实际的数字和有多少个交换了位置的数字即可。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <utility>
#include <map>
#include <string>
#include <cmath>
#include <vector>
#include <cstring>

using namespace std;

#define SQR(x) ((x)*(x))
#define LL long long
#define LOWBIT(x) ((x)&(-(x)))
#define PB push_back
#define MP make_pair

#define MAXN 211111

struct TreeArray{
    int tree[MAXN], n;
    void clear(int nn = MAXN - 10) {
        n = nn;
        memset(tree, 0, sizeof(tree[0]) * (n + 10));
    }
    void add(int x, int num = 1) {
        while (x <= n) {
            tree[x] += num;
            x += LOWBIT(x);
        }
    }
    int get(int x) {
        int res = 0;
        while (x > 0) {
            res += tree[x];
            x -= LOWBIT(x);
        }
        return res;
    }
} TA;

pair<int, int > query[MAXN];
vector <int > lsh;
vector <int > a;
int n;
LL ans;

void init() {
    cin >> n;
    int x, y;
    lsh.clear();
    for (int i = 0; i < n; ++i) {
        cin >> x >> y;
        query[i] = MP(x, y);
        lsh.PB(x); lsh.PB(y);
    }
    sort(lsh.begin(), lsh.end()); 
    lsh.erase(unique(lsh.begin(), lsh.end()), lsh.end());
}

#define INDEX(x) ((lower_bound(lsh.begin(), lsh.end(), (x))-lsh.begin())+1)

void solve() {
    ans = 0;
    a = lsh;
    TA.clear(lsh.size()+10);
    for (int i = 0; i < n; ++i) {
//        cout << query[i].first <<' '<< query[i].second << endl;
        swap(a[INDEX(query[i].first)-1], a[INDEX(query[i].second)-1]);
    }

//    for (int i = 0; i < lsh.size(); ++i) cout << lsh[i] <<' '; cout << endl;
//    for (int i = 0; i < a.size(); ++i) cout << a[i] <<' '; cout << endl;

    for (int i = (int)a.size() - 1; i >= 0; --i) {
        ans += TA.get(INDEX(a[i]) - 1);
        TA.add(INDEX(a[i]));
        int key = 0, t;
        t = INDEX(a[i]) - 1;
        if (i < t) {
            key = (lsh[t] - lsh[i] - 1) - (t - i - 1);
        } 
        else {
            key = (lsh[i] - lsh[t] - 1) - (i - t - 1);
        }
        ans += key;
//        cout << ans << endl;
    }
}

int main() {
//    freopen("test.txt", "r", stdin);
    ios::sync_with_stdio(false);
    init();
    solve();
    cout << ans << endl;
    return 0;
}