原题链接:http://poj.org/problem?id=2299
Ultra-QuickSort
Time Limit: 7000MS | Memory Limit: 65536K | |
Total Submissions: 31043 | Accepted: 11066 |
Description
![](http://poj.org/images/2299_1.jpg)
Ultra-QuickSort produces the output
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
Source
题意:求逆序数。
算法:归并排序,好像还有个树状数组的也可以解决,实在没弄明白,明天再想吧。。。
思路:归并合并时,一旦遇到前面一半的元素i大于后面一半中的元素j,那么就说明从素 i 开始一直到前面一半结束,每一个数都
和j组成逆序对,需交换。
注意:最坏的情况下 ans = 500000*500000/2要用long long型,开始没有注意WA了好久。
PS:应该算是很简单的一道题目了,KB神给我讲了一遍,还是看了好久才AC明天继续树状数组版本。。。
//Accepted 3692 KB 391 ms C++ 750 B 2013-03-08 21:37:57 #include<cstdio> #include<cstring> const int maxn = 500000 + 10; int a[maxn]; int t[maxn]; int n; __int64 ans; void merge_sort(int *a, int x, int y, int *t) { if(y-x > 1) { int m = x + (y-x)/2; int p = x, q = m, i = x; merge_sort(a, x, m, t); merge_sort(a, m, y, t); while(p < m && q < y) //合并 { if(a[p] <= a[q]) t[i++] = a[p++]; else { t[i++] = a[q++];//a[p]>a[q] ans += m-p; } } while(p < m) t[i++] = a[p++]; while(q < y) t[i++] = a[q++]; for(i = x; i < y; i++) a[i] = t[i]; } } int main() { while(scanf("%d", &n) != EOF) { ans = 0; if(n == 0) break; for(int i = 0; i < n; i++) scanf("%d", &a[i]); merge_sort(a, 0, n, t); printf("%I64d\n", ans); } return 0; }
利用树状数组离散化求逆序数,看了很久,还是无法完全理解,贴个KB神的用树状数组求解的代码吧。。。
http://www.cnblogs.com/kuangbin/archive/2012/08/09/2630042.html
/* POJ 2299 Ultra-QuickSort 求逆序数 离散化+树状数组 首先从小到大进行编号,从而实现离散化 然后利用树状数组来统计每个数前面比自己大的数的个数 */ #include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> using namespace std; const int MAXN=500010; int c[MAXN]; int b[MAXN]; int n; struct Node { int index;//序号 int v; }node[MAXN]; bool cmp(Node a,Node b) { return a.v<b.v; } int lowbit(int x) { return x&(-x); } void add(int i,int val) { while(i<=n) { c[i]+=val; i+=lowbit(i); } } int sum(int i) { int s=0; while(i>0) { s+=c[i]; i-=lowbit(i); } return s; } int main() { // freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); while(scanf("%d",&n),n) { for(int i=1;i<=n;i++) { scanf("%d",&node[i].v); node[i].index=i; } memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); //离散化 sort(node+1,node+n+1,cmp); //将最小的编号为1 b[node[1].index]=1; for(int i=2;i<=n;i++) { if(node[i].v!=node[i-1].v) b[node[i].index]=i; else b[node[i].index]=b[node[i-1].index]; } long long ans=0; //这里用的很好 //一开始c数组都是0,然后逐渐在b[i]处加上1; for(int i=1;i<=n;i++) { add(b[i],1); ans+=i-sum(b[i]); //统计每个数前面比自己大的数的个数 } printf("%I64d\n",ans); } return 0; }