poj_2299 Ultra-QuickSort（归并排序/树状数组 求逆序对）

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 57323 Accepted: 21183

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,
Ultra-QuickSort produces the output 
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

59105431230

Sample Output

60

利用归并排序的性质求逆序对，要注意最后输出值较大 要用long long。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define inf 0x3f3f3f3f#define maxn 500010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;int n;int a[maxn];int t[maxn];LL ans = 0;void Merge(int l, int m, int r){    int i = l, j = m+1, cot = l;    while(i <= m && j <= r)    {        if(a[i] > a[j]) t[cot++] = a[j++], ans += m - i + 1;        else t[cot++] = a[i++];    }    while(i <= m) t[cot++] = a[i++];    while(j <= r) t[cot++] = a[j++];    for(i = l; i <= r; i++) a[i] = t[i];}void Msort(int l, int r){    if(l < r)    {        int m = l + r >> 1;        Msort(l, m);        Msort(m+1, r);        Merge(l, m, r);    }}int main(){    while(scanf("%d", &n) && n)    {        ans = 0;        for(int i = 0; i < n; i++) scanf("%d", &a[i]);        Msort(0, n-1);        printf("%lld\n", ans);    }    return 0;}

也可以离散（也就是排序后用较小的数去代替原数）后用树状数组的性质做。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define inf 0x3f3f3f3f#define maxn 500010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;struct Node{    int v, pos;}a[maxn];int n;int aa[maxn]; //离散之后的数组int c[maxn];  //树状数组bool compare(Node a, Node b){    return a.v < b.v;}int lowbit(int i){    return i & (-i);}void update(int i, int val){    while(i <= n)    {        c[i] += val;        i += lowbit(i);    }}int Sum(int x){    int ans = 0;    while(x > 0)    {        ans += c[x];        x -= lowbit(x);    }    return ans;}int main(){    while(scanf("%d", &n) && n)    {        for(int i = 1; i <= n; i++)        {            scanf("%d", &a[i].v);            a[i].pos = i;        }        sort(a+1, a+n+1, compare);        for(int i = 1; i <= n; i++) aa[a[i].pos] = i;        memset(c, 0, sizeof(c));        LL ans = 0;        for(int i = 1; i <= n; i++)        {            update(aa[i], 1);            ans += i - Sum(aa[i]);        }        printf("%lld\n", ans);    }    return 0;}