[HDU] 1394 minimum inversion number-Brute Force reverse order, tree array reverse order, line segment tree reverse order, merge sort Reverse Order

I'm bored with it ..

In fact, the key to this question is not the method of reverse order, because it is also possible to use brute force.

To give a sequence with the length of n (n <= 5000), from 0 ~ N-1 digits. Each time we move the leftmost number to rightmost to form a new sequence. Then a total of N sequences can be formed. Find the minimum number of reverse orders in the N sequences.

I use array a [0 ~ N-1] to store raw data. You only need to find the sum of the reverse order of the original sequence, and then for each order of a [I] (0 <= I <n-1, use sum to subtract the number (that is, a [I]) of a smaller number on the right side of the object. after sum is used, the number (n-A [I]-1) on the left is shifted to the reverse order of a [I ].

1. Violence Law.

265 Ms

212 K

600 B

Problem: 1394 (minimum inversion number) Judge Status: Accepted
Runid: 2732590 language: C ++ Author: yueashuxia
Code render status: rendered by hdoj C ++ code render version 0.01 Beta

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;#define MAX 50001int n,a[MAX];int main (){    int i, j, MIN, sum ;    while(scanf("%d", &n) != EOF )    {        sum = 0 ;        for(i = 0; i < n; i ++)        {             scanf("%d", &a[i]);             for(j = 0; j < i; j ++)             {                   if(a[j] > a[i]) sum ++ ;              }        }        MIN = sum ;        for(i = 0; i < n-1; i ++){              sum = sum - a[i] + (n - a[i] - 1) ;              if(sum < MIN) MIN = sum ;        }        printf("%d/n", MIN);   }   return 0 ;}

 
2. Tree array method.

 
 

  
  
 
   
 
    
31 Ms
 
    
408 K
 
    
816 B
 
   
 
  
 
 
Problem: 1394 (minimum inversion number) Judge Status: Accepted
Runid: 2731410 language: C ++ Author: yueashuxia
Code render status: rendered by hdoj C ++ code render version 0.01 Beta
#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;#define MAX 50001int c[MAX],n,a[MAX];int getsum(int i){    int res;    for(res=0;i>0;i-=i&(-i))  res+=c[i];        return res;}void modify(int i,int m){    for(;i<=MAX;i+=i&(-i)) c[i]+=m;}int main (){    int i, j, MIN, sum ;    while(scanf("%d", &n) != EOF )    {        memset(c, 0, sizeof(c)) ;        sum = 0 ;        for(i = 0; i < n; i ++)        {             scanf("%d", &a[i]);        }        for(i = n-1; i >= 0; i --)        {             modify(a[i]+ 1, 1) ;             sum += getsum(a[i]) ;        }        MIN = sum ;        for(i = 0; i < n-1; i ++){              sum = sum - a[i] + (n - a[i] - 1) ;              if(sum < MIN) MIN = sum ;         }         printf("%d/n", MIN);   }   return 0 ;}
 
3. Line Segment tree method.

 
 

  
  
 
   
 
    
46 Ms
 
    
300 K
 
    
1639 B
 
   
 
  
 
 
Problem: 1394 (minimum inversion number) Judge Status: Accepted
Runid: 2733007 language: C ++ Author: yueashuxia
Code render status: rendered by hdoj C ++ code render version 0.01 Beta
#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;const int size = 50001 ;struct xds{    int lson, rson ;    int a, b, info ;}tree[size*3] ;int tot, n, a[size] ;void build(int a, int b){    int now = ++tot ;    tree[now].a = a ;    tree[now].b = b ;    tree[now].info = 0 ;    if(a == b) {        tree[now].lson = tree[now].rson = 0 ;           return ;    }    int mid = (a + b) >> 1 ;    tree[now].lson = tot + 1 ;    build(a, mid) ;    tree[now].rson = tot + 1 ;    build(mid + 1, b) ;}void update(int p, int a){    tree[p].info ++ ;    if(tree[p].a == tree[p].b) return ;    int mid = (tree[p].a + tree[p].b) >> 1 ;    if(a <= mid) update(tree[p].lson, a) ;    else update(tree[p].rson, a) ;}int query(int p, int a, int b){    if(tree[p].a == a && tree[p].b == b)    {         return tree[p].info ;     }     if(tree[p].a == tree[p].b) return tree[p].info;     int mid = (tree[p].a + tree[p].b) >> 1 ;     if(b <= mid) return query(tree[p].lson, a, b) ;     else if(a > mid) return query(tree[p].rson, a, b) ;     else return query(tree[p].lson, a, mid) + query(tree[p].rson, mid + 1, b) ;}int main (){    int i, j, MIN, sum ;    while(scanf("%d", &n) != EOF )    {        sum = tot = 0 ;        build(1, n) ;        for(i = 0; i < n; i ++)        {            scanf("%d", &a[i]);            sum += query(1, a[i]+2, n) ;            update(1, a[i]+ 1) ;        }        MIN = sum ;        for(i = 0; i < n-1; i ++){            sum = sum - a[i] + (n - a[i] - 1) ;             if(sum < MIN) MIN = sum ;           }        printf("%d/n", MIN);    }    return 0 ;}

  
4. Merge Sorting.

  
  

   
   
 
    
 
     
31 Ms
 
     
240 K
 
     
1169
B
 
    
 
   
  
  
Problem : 1394 ( Minimum Inversion Number )     Judge Status : AcceptedRunId : 2732771    Language : C++    Author : yueashuxiaCode Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta
# Include <stdio. h> # include <string. h >#include <iostream >#include <algorithm> using namespace STD; # define Max 50001 typedef int t; t num [Max], a [Max], sum, B [Max]; // sorts the num array in ascending order. A is the auxiliary array void mergesort (T left, t mid, t right) {t I = left, j = Mid + 1, K = 0; while (I <= Mid & J <= right) {If (Num [I] <= num [J]) A [k ++] = num [I ++]; else {sum + = mid-I + 1; // if this sentence is added and sum = 0 is initialized, the sum is the reverse number of the original sequence. A [k ++] = num [J ++];} while (I <= mid) A [k ++] = num [I + +]; While (j <= right) A [k ++] = num [J ++]; for (I = left, j = 0; I <= right; I ++) num [I] = A [J ++];} void Merge (T left, t right) {// grouping if (left <right) {Merge (left, (left + right)/2); merge (left + right)/2 + 1, right); mergesort (left, (left + right) /2, right) ;}} int main () {int N, I, j, Min, last; while (scanf ("% d", & N )! = EOF) {for (I = 0; I <n; I ++) {scanf ("% d", & num [I]); B [I] = num [I];} sum = 0; merge (0, n-1); min = sum; for (I = 0; I <n-1; I ++) {sum = sum-B [I] + (n-B [I]-1); If (sum <min) min = sum ;} printf ("% d/N", min);} return 0 ;}