CF540E

Codeforces 301 Div.2 E

题意是说,给一个按大小排序的正整数数列,给出n个对换操作,n<=10^5.问操作后该数列的逆序对对数.

思路跟普通的求逆序对对数思路是一样的,都是利用归并排序,只是原来的数这里用区间代替.应该说还是比较简单的.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include <bitset>
using namespace std;

#define sqr(x) (x)*(x)

typedef long long ll;

const ll mod=1000000007;

int n;
map<int,int> ma;
map<int,int>::iterator ite;
ll r=0;

struct inter
{
    int x,y;
    inter(int x=0,int y=0):x(x),y(y){}
};
bool operator <(inter a,inter b)
{
    return a.x<b.x;
}

inter p[1000005];

inter te[1000005];

void mergesort(int st,int en)//[st,en)
{
    if(en-st<=1) return;
    int mid=(st+en)/2;
    mergesort(st,mid);mergesort(mid,en);
    int i,j,tn;
    ll le=0;
    for(i=st;i<mid;i++) le=le+(p[i].y-p[i].x+1);
    i=st,j=mid,tn=0;
    while(i<mid&&j<en)
    {
        if(p[i]<p[j])
        {
            te[tn++]=p[i];
            le-=(p[i].y-p[i].x+1);
            ++i;
        }
        else
        {
            te[tn++]=p[j];
            r=r+le*(p[j].y-p[j].x+1);
            ++j;
        }
    }
    while(i<mid)
    {
        te[tn++]=p[i];++i;
    }
    while(j<en)
    {
        te[tn++]=p[j];++j;
    }
    for(i=st;i<en;i++) p[i]=te[i-st];
}

int main()
{
    int i,j;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        int x,y,xx,yy;
        scanf("%d %d",&x,&y);
        if(ma.find(x)!=ma.end()) xx=ma[x];
        else xx=x;
        if(ma.find(y)!=ma.end()) yy=ma[y];
        else yy=y;
        ma[x]=yy;ma[y]=xx;
        if(x==yy) ma.erase(x);
        if(y==xx) ma.erase(y);
    }
    n=0;
    int cur=1;
    for(ite=ma.begin();ite!=ma.end();ite++)
    {
        if(ite->first>cur) p[n++]=inter(cur,(ite->first)-1);
        p[n++]=inter(ite->second,ite->second);
        cur=ite->first+1;
    }
    mergesort(0,n);
    printf("%I64d\n",r);
    return 0;
}