POJ--2299 Ultra-QuickSort(离散化 + 求逆序数)

来源：互联网发布：罗马之子优化编辑：程序博客网时间：2024/05/16 06:04

传送门：Ultra-Quicksort

题意：求一个序列的逆序数。

分析：数据范围太大，需要先离散化处理。所谓离散化，就是缩小数与数之间的间隔，但要保证原来的大小关系不变。

然后是求逆序数，用归并排序的算法或者树状数组皆可。

-----1. 树状数组求逆序数

# include <iostream># include <cstdio># include <cstring># include <algorithm>using namespace std;//AC 407msconst int maxn = 500000 + 5;int a[maxn], c[maxn];int n;struct num{    int val, pos;    bool operator<(const num &a) const{        return val < a.val;    }}b[maxn];int lowbit(int x){    return x & -x;}void add(int x){    for(int i = x; i <= maxn; i += lowbit(i))        c[i]++;}int sum(int x){    int s = 0;    for(int i = x; i ; i -= lowbit(i))        s += c[i];    return s;}void solve(){    long long ans = 0;    for(int i = 1; i <= n; ++i)    {        add(a[i]);        ans += i - sum(a[i]);    }    cout << ans << endl;}void process()  //离散化处理{    for(int i = 1; i <= n; ++i)    {        scanf("%d", &b[i].val);        b[i].pos = i;    }    stable_sort(b + 1, b + n + 1);    for(int i = 1; i <= n; ++i)        a[b[i].pos] = i;}int main(){    while(cin >> n && n)    {        process();        memset(c, 0, sizeof(c));        solve();    }    return 0;}

-----2. 归并求逆序数

# include <iostream># include <cstdio># include <cstring># include <algorithm>using namespace std;//AC, 516msconst int maxn = 500000 + 5;int a[maxn], tmp[maxn];int n;long long ans;struct num{    int val, pos;    bool operator<(const num &a) const{        return val < a.val;    }}b[maxn];void Merge(int a[], int tmp[], int left, int mid, int right){    int i = left, j = mid + 1;    int k = left;    while(i <= mid && j <= right)    {        if(a[i] < a[j]){            tmp[k++] = a[i++];        }        else{            ans += mid - i + 1;            tmp[k++] = a[j++];        }    }    while(i <= mid)   tmp[k++] = a[i++];    while(j <= right) tmp[k++] = a[j++];    for(i = left; i <= right; ++i)        a[i] = tmp[i];}void MSort(int a[], int tmp[], int left, int right){    if(left < right)    {        int mid = (left + right) >> 1;        MSort(a, tmp, left, mid);        MSort(a, tmp, mid + 1, right);        Merge(a, tmp, left, mid, right);    }}void solve(){    MSort(a, tmp, 1, n);    cout << ans << endl;}void process(){    for(int i = 1; i <= n; ++i)    {        scanf("%d", &b[i].val);        b[i].pos = i;    }    stable_sort(b + 1, b + n + 1);    for(int i = 1; i <= n; ++i)        a[b[i].pos] = i;}int main(){    while(cin >> n && n)    {        ans = 0;        process();        solve();    }    return 0;}

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