Ultra-QuickSort
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 57461 | Accepted: 21231 |
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 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
哼你以为你让我WA我就不发日志了吗QAQ
经过快一年的努力终于改对了！！哈哈哈哈哈
问题出在离散化，相等的数我把弄了不同的位置所以错了
1 #include "bits/stdc++.h" 2 #define mem(a,b) memset(a,b,sizeof(a)) 3 using namespace std; 4 typedef long long LL; 5 const int MAX=500005; 6 int n; 7 struct Poi{ 8 int w,p,z; 9 }cc[MAX]; 10 int c[MAX]; 11 void add(int x,int y){for (;x<=n;x+=(x&-x)) c[x]+=y;} 12 int search(int x){int an(0);for (;x>0;x-=(x&-x)) an+=c[x];return an;} 13 bool cmp1(Poi x,Poi y){return x.w>y.w;} 14 bool cmp2(Poi x,Poi y){return x.p<y.p;} 15 void init(){ 16 int i,j; 17 for (i=1;i<=n;i++){ 18 scanf("%d",&cc[i].w); 19 cc[i].p=i; 20 } 21 sort(cc+1,cc+n+1,cmp1); 22 for (i=1;i<=n;i++){
if (vv[i].w!=vv[i-1].w)
vv[i].z=++tt;
else
vv[i].z=tt; 24 } 25 sort(cc+1,cc+n+1,cmp2); 26 mem(c,0); 27 } 28 int main(){ 29 freopen ("ultra.in","r",stdin); 30 freopen ("ultra.out","w",stdout); 31 int i,j; 32 int ans; 33 while (scanf("%d",&n),n){ 34 init();ans=0; 35 for (i=1;i<=n;i++){ 36 ans+=search(cc[i].z-1); 37 add(cc[i].z,1); 38 } 39 printf("%d\n",ans); 40 } 41 return 0; 42 }