求逆序数的两个方法

1.归并排序

#include <iostream>

using namespace std;

const int maxn = 1000;

int a[maxn],b[maxn];
int ans;

void merge_sort(int x,int y){
    if(y-x>1)
    {
        int m = x+(y-x)/2;
        int p = x, q = m,i = x;
        merge_sort(x,m);
        merge_sort(m,y);
        while(p<m||q<y){
            if(q>=y||(p<m&&a[p]<=a[q])) b[i++] = a[p++];
            else {b[i++]=a[q++];ans+=m-p;}
        }
        for(int i = x;i<y;i++) a[i] = b[i];
    }
}

int main()
{
    int n;
    while(cin>>n){
       for(int i=0;i<n;i++) cin>>a[i];
       ans=0;
       merge_sort(0,n);
       cout<<ans<<endl;
       for(int i=0;i<n;i++) cout<<a[i]<<" ";cout<<endl;
    }
    return 0;
}

2.离散+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 ≤ 109, ai ≠ 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

Note

In the first sample the sequence is being modified as follows: . It has 4 inversions formed by index pairs (1, 4), (2, 3), (2, 4) and (3, 4).

离散：

逆序数组：9 1  0 5 4 --> 5 2 1 4 3

这题有点特别：就是先把那些交换的点+那些中间的点（等价的几个，所以a[i].yy可能大于1）

注意：maxn要为4*1e5，因为不单只2倍的10的5次方操作!

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#include <cmath>
#include <stack>
#include <iomanip>
#include <map>

#define sc scanf
#define pf printf
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define CLR(x,m) memset(x,m,sizeof x);
#define LOCAL freopen("test.txt","r",stdin)
#define UCCU ios::sync_with_stdio(0);
#define XSD(i) cout << fixed << setprecision(i);
#define pii pair<int,int>
#define xx first
#define yy second

using namespace std;

typedef long long LL;
typedef double DB;

const double eps = 1e-6;
const int maxn = 400100;

map<int,int> mp,mp2;
map<int, int> :: iterator it;
pii a[maxn];
int b[maxn];
LL fen[maxn];

void add(int n, int k)
{
    for(; n < maxn; n += n & (-n))
      fen[n] += k;
}

int get(int n)
{
    int res = 0;
    for(; n > 0; n -= n & (-n))
      res += fen[n];
    return res;
}


void run()
{
    int n;
    cin>>n;
    REP(i,0,n){
        int u,v;cin>>u>>v;
        if(!mp[u]) mp[u]=u;
        if(!mp[v]) mp[v]=v;
        int temp = mp[u];
        mp[u] = mp[v];
        mp[v] = temp;
    }

    int pre=1,num=0;
    for(it = mp.begin();it!=mp.end();++it){
        int x = (*it).xx;int y = (*it).yy;
        if(x!=1&&x-pre-1){
            a[++num]=pii(x-1,x-pre-1);
            b[num]=x-1;
        }
        a[++num]=pii(y,1);
        b[num]=y;
        pre=x;
    }


    sort (b + 1, b + 1 + num);
    for (int i = 1; i <= num; i++) {
        mp2[b[i]] = i;
    }
    for (int i = 1; i <= num; i++) {
        a[i].xx = mp2[a[i].xx];
    }

    LL ans =0,tot=0;
    REP(i,1,num+1){
        ans+=1LL * (tot-get(a[i].xx))*a[i].yy;
        add(a[i].xx,a[i].yy);
        tot+=a[i].yy;
    }
    cout<<ans<<endl;
}

int main()
{
    //LOCAL;
    UCCU;//cin.tie();
    run();
    return 0;
}