POJ 2299 Ultra-QuickSort 离散化加树状数组求逆序对

http://poj.org/problem?id=2299

题意：求逆序对

题解：用树状数组。每读入一个数x,另a[x]=1.那么a数列的前缀和s[x]即为x前面（或者说，再x之前读入）小于x的个数，而逆序对就是x前面所有的数减去s[x]

　　关于离散化，由于5e5个数据是1e9范围的整数，上面的数组明显无法开到1e9，所以有一个离散化处理技巧：将每个数的下标记录下来，然后对原序列排序（下标也跟着排）。于是我们得到了一个下标的逆序对，观察发现其值等于原数列的逆序对。

ac代码：

坑：忘记初始化树状数组了。。。疯狂初始化别的数组orz

#define _CRT_SECURE_NO_WARNINGS
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<iomanip>
#include<cmath>
#include<cstdio>
#include<string>
#include<climits>
#include<stack>
#include<ctime>
#include<list>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<sstream>
#include<fstream>
#include<iostream>
#include<functional>
#include<algorithm>
#include<memory.h>
//#define INF LONG_MAX
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mp make_pair
#define pb push_back
#define mmm(a,b) memset(a,b,sizeof(a))
//std::ios::sync_with_stdio(false);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
void smain();
#define ONLINE_JUDGE
int main() {
    //ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    freopen("C:\Users\SuuTT\Desktop\test\in.txt", "r", stdin);
    freopen("C:\Users\SuuTT\Desktop\test\out.txt", "w", stdout);
    //ifstream in;
    //string filename;
    //getline(cin, filename,'
');
    //in.open(filename);

    long _begin_time = clock();
#endif
    smain();
#ifndef ONLINE_JUDGE
    long _end_time = clock();
    printf("time = %ld ms.", _end_time - _begin_time);
#endif
    return 0;
}
int dir[4][2] = { 1,0,0,1,-1,0,0,-1 };
const int maxn = 5e5 + 5;
int n, m;
ll a[maxn];
pair<int, int> aa[maxn];
int d[maxn];
int level[maxn];
int lowbit(int x) { return x & (-x); }
void add(int x, int v) {//a[x]+=v;
    while (x <= maxn) {
        d[x] += v;
        x += lowbit(x);
    }

}
int query(int x) {
    int res = 0;
    while (x) {
        res += d[x];
        x -= lowbit(x);
    }
    return res;
}
struct node {
    
    int x, y;
    node(int x = 0, int y = 0) :x(x), y(y) {}


};

void Run() {

}

void smain() {
    while (cin >> n){
        if (!n)return;

        mmm(d, 0);
        rep(i, 1, n) {

            scanf("%d", &aa[i].first);
            aa[i].second = i;
        }
            sort(aa+1, aa + n+1);
            rep(i, 1, n)a[i] = aa[i].second;//a[aa[i].second]=i;

            ll ans = 0;
            rep(i, 1, n) {
                add(a[i], 1);
                ans += (i - query(a[i]));
                
            }
            cout << ans << endl;
    }
    
}