[树状数组]poj 2299

题意

????? 求一列数字的逆序数。

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思路

????? 由于数字很大，所以不能直接求，这里先对原数列排序之后，求出原下标的逆序数即可

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define lowbit(x) (x&(-x))
using namespace std;
const int nMax = 500010;
typedef long long ll;

struct rrr{
    ll a,b;
}num[nMax];

bool cmp(rrr a, rrr b){
    if(a.a<b.a)return 1;
    return 0;
}
int n, arr[nMax];

void add(int loc, int val){
    while(loc <= n){
        arr[loc] += val;
        loc += lowbit(loc);
    }
}

int sum(int loc){
    int res = 0;
    while(loc >= 1){
        res += arr[loc];
        loc -= lowbit(loc);
    }
    return res;
}

int main(){
    int i;
    while(scanf("%d",&n)!=EOF&&n){
        for(i = 1; i <= n; i ++){
            scanf("%I64d",&num[i].a);
            num[i].b = i;
        }
        sort(num + 1, num + n + 1 ,cmp);
        memset(arr, 0,sizeof(arr));
        ll res = 0;
        for(i = 1; i <= n; i++){
            res += num[i].b - 1 - sum(num[i].b);
//            cout<<num[i].b<<" "<<sum(num[i].b)<<endl;
            add(num[i].b, 1);
        }
        printf("%I64d\n",res);
    }
    return 0;
}

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[树状数组]poj 2299

原文：http://bbezxcy.iteye.com/blog/2166082