HDU 3015 Disharmony Trees 【 树状数组 】

题意：给出n棵树，给出横坐标x，还有它们的高度h，先按照横坐标排序，则它们的横坐标记为xx,

再按照它们的高度排序，记为hh

两颗树的差异度为 abs(xx[i] - xx[j]) * min(hh[i],hh[j])，求所有的差异度的和

和上面一道题一样，只不过这题是要Min的值，就将h从大到小排序，保证每一个h都是当前最小的

然后维护比当前x小的坐标的个数，当前区间的总和，前面比x小的坐标的和

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include <cmath>
 #include<stack>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<algorithm>
 using namespace std;

 typedef long long LL;
 const int INF = (<<)-;
 const int mod=;
 const int maxn=;

 int n;
 struct node{ int x,xx,h,hh,id;};

 node a[maxn],b[maxn],q[maxn];
 int c[maxn],d[maxn];
 int s[maxn];

 int cmp1(node n1,node n2){ return n1.x < n2.x; }
 int cmp2(node n1,node n2){ return n1.h < n2.h;}
 int cmp3(node n1,node n2){ return n1.h > n2.h;}

 int lowbit(int x){ return x & (-x);}

 LL sum(int c[],int x){
     LL ret=;
     while(x>){
         ret+=c[x]; x-=lowbit(x);
     }
     return ret;
 }

 void add(int c[],int x,int d){
     while(x < maxn){
         c[x]+=d;x+=lowbit(x);
     }
 }

 int main(){
     while(scanf("%d",&n) != EOF){
         memset(c,,sizeof(c));
         memset(d,,sizeof(d));
         for(int i=;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].h);

         sort(a+,a+n+,cmp1);a[].xx = ;
         for(int i=;i<=n;i++){
             if(a[i].x == a[i-].x) a[i].xx = a[i-].xx;
             else a[i].xx = i;
         }
         sort(a+,a+n+,cmp2);a[].hh = ;
         for(int i=;i<=n;i++){
             if(a[i].h != a[i-].h) a[i].hh =i;
             else a[i].hh = a[i-].hh;
         }
         sort(a+,a+n+,cmp3);

         //for(int i=1;i<=n;i++)
         //printf("a[%d].x = %d a[%d].h = %d\n",i,a[i].xx,i,a[i].hh);

         LL ans = ;
         for(int i=;i<=n;i++){
             LL x = sum(c,a[i].xx);
             LL totalfront = sum(d,a[i].xx);
             LL total = sum(d,maxn);
             ans += a[i].hh * (x*a[i].xx - totalfront + total - totalfront - (i-x-)*a[i].xx);
         //    printf("x= %I64d  totalfront=%I64d total=%I64d  ans = %I64d\n",x,totalfront,total,ans);
             add(c,a[i].xx,);
             add(d,a[i].xx,a[i].xx);
         }
         printf("%I64d\n",ans);
     }
     return ;
 }