xtu数据结构 C. Ultra-QuickSort

C. Ultra-QuickSort

Time Limit: 7000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

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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

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

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #define LL long long
 #define INF 0x3f3f3f
 using namespace std;
 const int maxn = ;
 struct node {
     int val,index;
 } p[maxn];
 LL tree[maxn];
 bool cmp(const node &x,const node &y) {
     return x.val > y.val;
 }
 int lowbit(int x) {
     return x&(-x);
 }
 void update(int x,int val) {
     for(int i = x; i < maxn; i += lowbit(i)) {
         tree[i] += val;
     }
 }
 LL sum(int x) {
     LL ans = ;
     for(int i = x; i; i -= lowbit(i)) {
         ans += tree[i];
     }
     return ans;
 }
 int main() {
     int n,i;
     LL ans;
     while(scanf("%d",&n),n) {
         for(i = ; i < n; i++) {
             scanf("%d",&p[i].val);
             p[i].index = i+;
         }
         sort(p,p+n,cmp);
         memset(tree,,sizeof(tree));
         int pre = INT_MIN;
         for(ans = i = ; i < n; i++) {
             update(p[i].index,);
             ans += sum(p[i].index-);

         }
         printf("%lld\n",ans);
     }
     return ;
 }

归并排序

 #include <cstdio>
 #define LL long long
 LL sum,dt[];
 void mysort(int lft,int rht,int step){
     static LL temp[];
     int md = lft + (step>>),i = ,k = ,j = ;
     while(lft + i < md && md + j < rht){
         if(dt[lft+i] > dt[md+j]){temp[k++] = dt[md+j];j++;
         }else{temp[k++] = dt[lft+i];i++;sum += j;}
     }
     while(lft+i < md){temp[k++] = dt[lft+i];i++;sum += j;}
     while(md+j < rht){temp[k++] = dt[md+j];j++;}
     for(i = ; i < k; i++) dt[lft+i] = temp[i];
 }
 void ms(int n){
     int len = ,step = ,m,i,u,v;
     sum = ;
     while(len < n){len <<= ;}
     m = len/step;
     while(step <= len){
         for(i = ; i < m; i++){
             u = i*step;v = (i+)*step;
             mysort(u,v>n?n:v,step);
         }
         step <<= ;m = len/step;
     }
 }
 int main(){
     int n,i;
     while(scanf("%d",&n),n){
         for(i = ; i < n; i++)
             scanf("%d",dt+i);
         ms(n);
         printf("%lld\n",sum);
     }
     return ;
 }