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;}