CF 61E 树状数组+离散化 求逆序数加强版 三个数逆序

http://codeforces.com/problemset/problem/61/E

题意是求 i<j<k && a[i]>a[j]>a[k] 的对数

会树状数组求逆序数的话,这个推一下就能出结果:
做法:
1、离散化,由于a[i]能够达到1e9

2、插入a[i]的时候,记录x[i]=i-sum(a[i]); a[i]之前比a[i]大的有x[i]个

3、插入完毕后,求a[i] 之后比a[i]小的数的个数y[i]

ans=segma(x[i]*y[i])   注意x[i]*y[i]会超出int  由于这wa了一次


#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const int INF = 100000000;
const int MAXN  = 1e6+100;
ll x[MAXN],y[MAXN];
int a[MAXN],tmp[MAXN];
ll c[MAXN];
int n;

inline int lowb(int x) {return x&(-x);}

void update(int x, int d)
{
    while(x<=n)
    {
        c[x]+=d;
        x+=lowb(x);
    }
}

ll sum(int x)
{
    ll ret=0;
    while(x>0)
    {
        ret+=c[x];
        x-=lowb(x);
    }
    return ret;
}

bool cmp(const int i, const int j)
{
    return a[i]<a[j];
}


void dis()
{
    sort(tmp+1,tmp+1+n,cmp);
    int tt=0,pre=a[tmp[1]]-1;
    for(int i=1;i<=n;i++)
    {
        if(pre!=a[tmp[i]])
        {
            pre=a[tmp[i]];
            a[tmp[i]]=++tt;
        }
        else
            a[tmp[i]]=tt;
    }
}

int main()
{
    //IN("B.txt");
    ll ans=0;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            tmp[i]=i;
        }
        dis();
        /////////////////
        /*for(int i=1;i<=n;i++)
        {
            printf("a[%d]=%d\n",i,a[i]);
        }*/
        /////////////////
        CL(c,0);
        ans=0;
        for(int i=1;i<=n;i++)
        {
            update(a[i],1);
            x[i]=i-sum(a[i]);//a[i]之前比a[i]大的有`x[i]个
            y[i]=sum(a[i]-1);//a[i]之前比a[i]小的数
        }
        for(int i=1;i<=n;i++)
        {
            y[i]=sum(a[i]-1)-y[i];//a[i] 之后比a[i]小的数的个数
            ans+=x[i]*y[i];
        }
        /////////////////
       /* for(int i=1;i<=n;i++)
            printf("i=%d x=%d y=%d\n",i,x[i],y[i]);*/
        /////////////////
        printf("%I64d\n",ans);
    }

    return 0;
}



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