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 = 500002;
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 = 0;
    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 = 0; i < n; i++) {
            scanf("%d",&p[i].val);
            p[i].index = i+1;
        }
        sort(p,p+n,cmp);
        memset(tree,0,sizeof(tree));
        int pre = INT_MIN;
        for(ans = i = 0; i < n; i++) {
            update(p[i].index,1);
            ans += sum(p[i].index-1);

        }
        printf("%lld
",ans);
    }
    return 0;
}

归并排序

#include <cstdio>
#define LL long long
LL sum,dt[500010];
void mysort(int lft,int rht,int step){
    static LL temp[500010];
    int md = lft + (step>>1),i = 0,k = 0,j = 0;
    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 = 0; i < k; i++) dt[lft+i] = temp[i];
}
void ms(int n){
    int len = 1,step = 2,m,i,u,v;
    sum = 0;
    while(len < n){len <<= 1;}
    m = len/step;
    while(step <= len){
        for(i = 0; i < m; i++){
            u = i*step;v = (i+1)*step;
            mysort(u,v>n?n:v,step);
        }
        step <<= 1;m = len/step;
    }
}
int main(){
    int n,i;
    while(scanf("%d",&n),n){
        for(i = 0; i < n; i++)
            scanf("%d",dt+i);
        ms(n);
        printf("%lld
",sum);
    }
    return 0;
}