hdu 3015 Disharmony Trees (离散化+树状数组)

Disharmony Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 663    Accepted Submission(s): 307

Problem Description

One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them.

She finds that trees maybe disharmony and the Disharmony Value between two trees is associated with two value called FAR and SHORT.

The FAR is defined as the following:If we rank all these trees according to their X Coordinates in ascending order.The tree with smallest X Coordinate is ranked 1th.The trees with the same X Coordinates are ranked the same. For example,if there are 5 tree with X Coordinates 3,3,1,3,4. Then their ranks may be 2,2,1,2,5. The FAR of two trees with X Coordinate ranks D1 and D2 is defined as F = abs(D1-D2).

The SHORT is defined similar to the FAR. If we rank all these trees according to their heights in ascending order，the tree with shortest height is ranked 1th.The trees with the same heights are ranked the same. For example, if there are 5 tree with heights 4,1,9,7,4. Then their ranks may be 2,1,5,4,2. The SHORT of two trees with height ranks H1 and H2 is defined as S=min(H1,H2).

Two tree’s Disharmony Value is defined as F*S. So from the definition above we can see that, if two trees’s FAR is larger , the Disharmony Value is bigger. And the Disharmony value is also associated with the shorter one of the two trees.

Now give you every tree’s X Coordinate and their height , Please tell Sophia the sum of every two trees’s Disharmony value among all trees.

 

Input

There are several test cases in the input

For each test case, the first line contain one integer N (2 <= N <= 100,000) N represents the number of trees.

Then following N lines, each line contain two integers : X, H (0 < X,H <=1,000,000,000 ), indicating the tree is located in Coordinates X and its height is H.

 

Output

For each test case output the sum of every two trees’s Disharmony value among all trees. The answer is within signed 64-bit integer.

 

Sample Input

2

10 100

20 200

4

10 100

50 500

20 200

20 100

 

Sample Output

1

13

 

Source

2009 Multi-University Training Contest 12 - Host by FZU

 

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 //203MS    4548K    1948 B    C++
 /*
     题意：
         给出n个树的位置和高度，求每两棵数的位置排名差绝对值与高度排名较小的的积的和。

     离散化+树状数组:
         不错的一道题，我卡在结果的计算，的确对树状数组计算的能力的理解不够！
         用两个排序排出位置和高度的排名，然后再对高度排名在进行计算，计算公式：
             ans+=p[i].hid*(cnt*p[i].xid-sum+tsum-sum-(i-cnt-1)*p[i].xid); //每个高度
     算法时间复杂度为O(n*lgn)

 */
 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #define N 100005
 struct node{
     int x,h;
     int xid,hid;
     int id;
 }p[N],p1[N],p2[N];
 __int64 cnum[N],clen[N];
 int cmpx(const void *a,const void*b)
 {
     return (*(node*)a).x-(*(node*)b).x;
 }
 int cmph(const void*a,const void*b)
 {
     return (*(node*)a).h-(*(node*)b).h;
 }
 int cmp0(const void*a,const void*b)
 {
     return (*(node*)b).hid-(*(node*)a).hid;
 }
 int lowbit(int k)
 {
     return (-k)&k;
 }
 void update(__int64 c[],int k,int detal)
 {
     for(;k<N;k+=lowbit(k))
         c[k]+=detal;
 }
 __int64 getsum(__int64 c[],int k)
 {
     __int64 s=;
     for(;k>;k-=lowbit(k))
         s+=c[k];
     return s;
 }
 int main(void)
 {
     int n,x,h;
     while(scanf("%d",&n)!=EOF)
     {
         memset(cnum,,sizeof(cnum));
         memset(clen,,sizeof(clen));
         for(int i=;i<=n;i++){
             scanf("%d%d",&p[i].x,&p[i].h);
             p[i].id=i;
             p1[i]=p2[i]=p[i];
         }
         qsort(p1+,n,sizeof(p1[]),cmpx);
         for(int i=,j=;i<=n;i++){
             if(i> && p1[i].x!=p1[i-].x) j=i;
             p[p1[i].id].xid=j;
         }
         qsort(p2+,n,sizeof(p2[]),cmph);
         for(int i=,j=;i<=n;i++){
             if(i> && p2[i].h!=p2[i-].h) j=i;
             p[p2[i].id].hid=j;
         }
         qsort(p+,n,sizeof(p[]),cmp0);
         __int64 ans=;
         //for(int i=1;i<=n;i++) printf("*%d %d\n",p[i].hid,p[i].xid);
         update(cnum,p[].xid,);
         update(clen,p[].xid,p[].xid);
         for(int i=;i<=n;i++){
             __int64 cnt=getsum(cnum,p[i].xid);
             __int64 sum=getsum(clen,p[i].xid);
             __int64 tsum=getsum(clen,N-);
             ans+=p[i].hid*(cnt*p[i].xid-sum+tsum-sum-(i-cnt-)*p[i].xid);
             update(cnum,p[i].xid,);
             update(clen,p[i].xid,p[i].xid);
         }
         printf("%I64d\n",ans);
     }
     return ;
 }