題干：

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

5 9 1 0 5 4 3 1 2 3 0

Sample Output

6 0

題目大意：

大概題意就是給你n個(gè)數(shù)，讓你求逆序?qū)Φ膫€(gè)數(shù)。

解題報(bào)告：

? ? 這題是求逆序?qū)Φ膫€(gè)數(shù)，直接上權(quán)值樹(shù)狀數(shù)組，然后求個(gè)逆序數(shù)就可以了，這里提供兩種思路，一種是枚舉每一個(gè)數(shù)然后看截止當(dāng)前，他前面出現(xiàn)過(guò)多少比他大的（查后綴1的個(gè)數(shù)）。另一種是看截止當(dāng)前，他后面有多少比他小的，也就是還有多少比他小的還沒(méi)出現(xiàn)（查前綴0的個(gè)數(shù)）。

在此附上對(duì)應(yīng)兩種方法的AC代碼：

AC代碼1：

#include<bits/stdc++.h> #define ll long long using namespace std; const int MAX = 500005 ; int a[500005],b[MAX],c[500005]; int n; int lowbit(int x){return x&-x;} void update(int x,int val) {while(x<MAX) {c[x] += val;x+=lowbit(x);} } int query(int x) {int res = 0;while(x>0) {res += c[x];x-=lowbit(x);}return res; } int main() {while(~scanf("%d",&n)) {if(n==0) break;ll ans = 0;memset(c,0,sizeof c);for(int i = 1; i<=n; i++) scanf("%d",a+i),b[i] = a[i];sort(b+1,b+n+1);int top = unique(b+1,b+n+1) - b - 1;for(int i = 1; i<=n; i++) {int pos = lower_bound(b+1,b+top+1,a[i]) - b;update(pos,1);ans += query(MAX-1) - query(pos);//ans += (pos-1)-query(pos-1); // printf("yy\n");}printf("%lld\n",ans);}return 0 ;}

總結(jié)：

? ?1.注意update和query中while都不能寫等號(hào)。查詢query的時(shí)候越界問(wèn)題，所以需要查詢MAX-1；

AC代碼2：

//沒(méi)寫n==0的break？ #include<bits/stdc++.h> #define ll long long using namespace std; const int MAX = 500005 ; int a[500005],b[MAX],c[500005]; int n; int lowbit(int x){return x&-x;} void update(int x,int val) {while(x<=MAX) {c[x] += val;x+=lowbit(x);} } int query(int x) {int res = 0;while(x>0) {res += c[x];x-=lowbit(x);}return res; } int main() {while(~scanf("%d",&n)) {if(n==0) break;ll ans = 0;memset(c,0,sizeof c);for(int i = 1; i<=n; i++) scanf("%d",a+i),b[i] = a[i];sort(b+1,b+n+1);int top = unique(b+1,b+n+1) - b - 1;for(int i = 1; i<=n; i++) {int pos = lower_bound(b+1,b+top+1,a[i]) - b;update(pos,1);ans += (pos-1)-query(pos-1);}printf("%lld\n",ans);}return 0 ;}

?

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