标签:style blog http java color 使用
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.
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.
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.
5 9 1 0 5 4 3 1 2 3 0
6 0
解题:求逆序数,归并排序或者快排+树状数组都可以。坑爹的地方在于要使用long long ,害我WA了几次。逗比。。。。。。
树状数组+快速排序
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #define LL long long 13 #define INF 0x3f3f3f 14 using namespace std; 15 const int maxn = 500002; 16 struct node { 17 int val,index; 18 } p[maxn]; 19 LL tree[maxn]; 20 bool cmp(const node &x,const node &y) { 21 return x.val > y.val; 22 } 23 int lowbit(int x) { 24 return x&(-x); 25 } 26 void update(int x,int val) { 27 for(int i = x; i < maxn; i += lowbit(i)) { 28 tree[i] += val; 29 } 30 } 31 LL sum(int x) { 32 LL ans = 0; 33 for(int i = x; i; i -= lowbit(i)) { 34 ans += tree[i]; 35 } 36 return ans; 37 } 38 int main() { 39 int n,i; 40 LL ans; 41 while(scanf("%d",&n),n) { 42 for(i = 0; i < n; i++) { 43 scanf("%d",&p[i].val); 44 p[i].index = i+1; 45 } 46 sort(p,p+n,cmp); 47 memset(tree,0,sizeof(tree)); 48 int pre = INT_MIN; 49 for(ans = i = 0; i < n; i++) { 50 update(p[i].index,1); 51 ans += sum(p[i].index-1); 52 53 } 54 printf("%lld\n",ans); 55 } 56 return 0; 57 }
归并排序
1 #include <cstdio> 2 #define LL long long 3 LL sum,dt[500010]; 4 void mysort(int lft,int rht,int step){ 5 static LL temp[500010]; 6 int md = lft + (step>>1),i = 0,k = 0,j = 0; 7 while(lft + i < md && md + j < rht){ 8 if(dt[lft+i] > dt[md+j]){temp[k++] = dt[md+j];j++; 9 }else{temp[k++] = dt[lft+i];i++;sum += j;} 10 } 11 while(lft+i < md){temp[k++] = dt[lft+i];i++;sum += j;} 12 while(md+j < rht){temp[k++] = dt[md+j];j++;} 13 for(i = 0; i < k; i++) dt[lft+i] = temp[i]; 14 } 15 void ms(int n){ 16 int len = 1,step = 2,m,i,u,v; 17 sum = 0; 18 while(len < n){len <<= 1;} 19 m = len/step; 20 while(step <= len){ 21 for(i = 0; i < m; i++){ 22 u = i*step;v = (i+1)*step; 23 mysort(u,v>n?n:v,step); 24 } 25 step <<= 1;m = len/step; 26 } 27 } 28 int main(){ 29 int n,i; 30 while(scanf("%d",&n),n){ 31 for(i = 0; i < n; i++) 32 scanf("%d",dt+i); 33 ms(n); 34 printf("%lld\n",sum); 35 } 36 return 0; 37 }
xtu数据结构 C. Ultra-QuickSort,布布扣,bubuko.com
标签:style blog http java color 使用
原文地址:http://www.cnblogs.com/crackpotisback/p/3857991.html