SGU 180 Inversions (树状数组+离散化)

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Inversions
Time Limit: 250MS Memory Limit: 4096KB 64bit IO Format: %I64d & %I64u
Submit

Status

Description
180. Inversions

time limit per test: 0.25 sec.
memory limit per test: 4096 KB
input: standard
output: standard

There are N integers (1<=N<=65537) A1, A2,.. AN (0<=Ai<=10^9). You need to find amount of such pairs (i, j) that 1<=i< j<=N and A[i]>A[j].

Input
The first line of the input contains the number N. The second line contains N numbers A1…AN.

Output
Write amount of such pairs.

Sample test(s)

Input
5
2 3 1 5 4

Output
3

Author: Stanislav Angelyuk
Resource: Saratov ST team Spring Contest #1
Date: 18.05.2003

水树状数组

#include <cstring>#include <cstdio>#include <iostream>#include <string.h>#include <algorithm>#include <vector>using namespace std;int n;int num[70000];int tree[70000*4];vector <int> list;int cnt=0;int lowbit(int x){    return x&(-x);}void update(int x){    for(int i=x;i<=cnt;i+=lowbit(i))        tree[i]++;}int sum(int x){    int ans=0;    for(int i=x;i>0;i-=lowbit(i))        ans+=tree[i];    return ans;}int main(){    while(~scanf("%d",&n))    {        list.clear();        memset(tree,0,sizeof(tree));        for(int i=1;i<=n;i++)        {            scanf("%d",&num[i]);            list.push_back(num[i]);        }        sort(list.begin(),list.end());        list.erase(unique(list.begin(),list.end()),list.end());        cnt=list.size();        long long ans=0;        for(int i=n;i>=1;i--)        {            int pos=lower_bound(list.begin(),list.end(),num[i])-list.begin()+1;            update(pos);            ans+=sum(pos-1);        }        printf("%lld\n",ans);    }}

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