离散化的写法

example　　POJ 2229

Ultra-QuickSort

Time Limit: 7000MS　　Memory Limit: 65536K

 

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
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

Source

Waterloo local 2005.02.05

---------------------------------------------

话说这题配的图何解？

--------------------------------------------

 

Solution

离散化+树状数组

---------------------

1.　　map

TLE的姿势

#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
using namespace std;

const int N(5e5+5);
int bit[N];
int a[N], b[N];
long long ans;
map<int,int> mp;
int sum(int x){
    int res=0;
    for(; x; res+=bit[x], x-=x&-x);
    return res;
}
void add(int x, int n){
    for(; x<=n; bit[x]++, x+=x&-x);
}
int main(){
    for(int n; scanf("%d", &n), n;){
        for(int i=0; i<n; i++){
            scanf("%d", a+i);
            b[i]=a[i];
        }
        sort(b, b+n);
        mp.clear();
        int tot=0;
        for(int i=0; i<n; i++)
            if(!mp[b[i]]) mp[b[i]]=++tot;
        ans=0;
        memset(bit, 0, sizeof(bit));
        for(int i=0, id; i<n; i++){
            id=mp[a[i]];
            ans+=sum(n)-sum(id);
            add(id, tot);
        }
        printf("%lld
", ans);
    } 
}

注意加粗的几行，其实没必要排序，map本身就有序。

 ans+=sum(n)-sum(id);

这一句还可以优化

ans+=i-sum(id-1);

AC的姿势

#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
using namespace std;

const int N(5e5+5);
int bit[N], a[N];
long long ans;
map<int,int> mp;
int sum(int x){
    int res=0;
    for(; x; res+=bit[x], x-=x&-x);
    return res;
}
void add(int x, int n){
    for(; x<=n; bit[x]++, x+=x&-x);
}
int main(){
    for(int n; scanf("%d", &n), n; mp.clear()){
        for(int i=0; i<n; i++){
            scanf("%d", a+i);
            mp[a[i]];
        }
        int tot=0;
        for(map<int,int>::iterator it=mp.begin(); it!=mp.end(); it++)
            it->second=++tot;
        ans=0;
        memset(bit, 0, sizeof(bit));
        for(int i=0, id; i<n; i++){
            id=mp[a[i]];
            ans+=i-sum(id-1);
            add(id, tot);
        }
        printf("%lld
", ans);
    } 
}

这样常数还是很大，3766 MS。

-----------------------

2.　　间接排序

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int N(5e5+5);
int bit[N], a[N], id[N], Rank[N];
long long ans;
int sum(int x){
    int res=0;
    for(; x; res+=bit[x], x-=x&-x);
    return res;
}
void add(int x, int n){
    for(; x<=n; bit[x]++, x+=x&-x);
}
bool cmp(const int &i, const int &j){
    return a[i]<a[j];
}
int main(){
    for(int n; scanf("%d", &n), n;){
        for(int i=0; i<n; i++)
            scanf("%d", a+i);

        for(int i=0; i<n; i++)
            id[i]=i;
        sort(id, id+n, cmp);        
        for(int i=0; i<n; i++)
            Rank[id[i]]=i+1;

        ans=0;
        memset(bit, 0, sizeof(bit));
        for(int i=0; i<n; i++){
            ans+=i-sum(Rank[i]-1);
            add(Rank[i], n);
        }
        printf("%lld
", ans);
    } 
}

代码中，id[i]表示第i小的数的下标，Rank[i]表示a[i]是第几小的（即a[i]的Rank）。

注意：题目已明确输入是"a sequence of n distinct integers"

对于一般情况

        for(int i=0; i<n; i++)
            Rank[id[i]]=i+1;

应改成

        Rank[id[0]]=1;        
        for(int i=1; i<n; ){
            Rank[id[i]]=Rank[id[i-1]]+1;
            for(i++; i<n&&a[id[i]]==a[id[i-1]]; i++)
                Rank[id[i]]=Rank[id[i-1]];
        }

547 MS

--------------------------------------

3.　　unique + 二分查找

当然，这道题所有数都不相同，不必unique，代码中给出的是针对一般情况的写法

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N(5e5+5);
int bit[N], a[N], b[N];
long long ans;
int sum(int x){
    int res=0;
    for(; x; res+=bit[x], x-=x&-x);
    return res;
}
void add(int x, int n){
    for(; x<=n; bit[x]++, x+=x&-x);
}
int main(){
    for(int n; scanf("%d", &n), n;){
        for(int i=0, v; i<n; i++){
            scanf("%d", a+i);
            b[i]=a[i];
        }
        sort(b, b+n);
        int *End=unique(b, b+n);
        ans=0;
        memset(bit, 0, sizeof(bit));
        for(int i=0, id; i<n; i++){
            id=lower_bound(b, End, a[i])-b;
            ans+=i-sum(id);
            add(id+1, n);
        }
        printf("%lld
", ans);
    } 
}

688 MS

注意：

id=lower_bound(b, End, a[i])-b;

不能写成

id=find(b, b+n, a[i])-b;

find()的复杂度是O(n)

-------------------------------------

这道题还有一种更为巧妙（且美妙）的解法，不必离散化，请读者细细品味

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N(5e5+5);
int bit[N];
struct Node{
    int x, y;
}a[N];
long long ans;
int sum(int x){
    int res=0;
    for(; x; res+=bit[x], x-=x&-x);
    return res;
}
void add(int x, int n){
    for(; x<=n; bit[x]++, x+=x&-x);
}
bool cmp(const Node &a, const Node &b){
    return a.x > b.x;
}
int main(){
    for(int n; scanf("%d", &n), n;){
        for(int i=0, v; i<n; i++){
            scanf("%d", &v);
            a[i]={v, i+1};
        }
        sort(a, a+n, cmp);
        ans=0;
        memset(bit, 0, sizeof(bit));
        for(int i=0; i<n; i++){
            ans+=sum(a[i].y);
            add(a[i].y, n);
        }
        printf("%lld
", ans);
    } 
}

这种做法通常称为在线转离线

375 MS