POJ - 2299 - Ultra-QuickSort = 归并排序 + 逆序对 / 树状数组

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

求逆序对最简单的绝对不会是树状数组，一定是归并排序（认真），不过树状数组会不会快一点呢？理论上应该是树状数组快一点（假如不进行离散化）。

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;

const int MAXN = 500000;
int a[MAXN + 5];
int tmp[MAXN + 5];

//分治，对[l,r]闭区间进行归并排序并统计逆序数
ll merge(int l, int r) {
    if(l == r)
        return 0;
    int m = (l + r) >> 1;
    ll res = 0;
    res += merge(l, m);
    res += merge(m + 1, r);
    int i = l, j = m + 1, k = 0;
    while(i <= m || j <= r) {
        if(i > m) {
            tmp[++k] = a[j++];
        } else if(j > r) {
            tmp[++k] = a[i++];
        } else {
            if(a[i] <= a[j]) {
                tmp[++k] = a[i++];
            } else {
                res += m + 1 - i;
                tmp[++k] = a[j++];
            }
        }
    }
    k = 0;
    for(int i = l; i <= r; ++i) {
        a[i] = tmp[++k];
    }
    return res;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int n;
    while(~scanf("%d", &n)) {
        if(n == 0)
            break;
        for(int i = 1; i <= n; ++i)
            scanf("%d", &a[i]);
        printf("%lld\n", merge(1, n));
    }
}

带离散化的树状数组果然常数大得不行。

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;

const int MAXN = 500000;
int a[MAXN + 5];
int x[MAXN + 5];
int bit[MAXN + 5];

int n;

void add(int x) {
    for(; x <= n; x += (x & -x))
        ++bit[x];
}

ll sum(int x) {
    ll res = 0;
    for(; x; x -= (x & -x))
        res += bit[x];
    return res;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    while(~scanf("%d", &n)) {
        if(n == 0)
            break;
        memset(bit, 0, sizeof(bit[0]) * (n + 1));
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
            x[i] = a[i];
        }
        sort(x + 1, x + 1 + n);
        int xt = unique(x + 1, x + 1 + n) - (x + 1);
        ll res = 0;
        for(int i = 1; i <= n; ++i) {
            a[i] = lower_bound(x + 1, x + 1 + xt, a[i]) - x;
            add(a[i]);
            res += i - sum(a[i]);
        }
        printf("%lld\n", res);
    }
}

原文地址：https://www.cnblogs.com/Inko/p/11729741.html