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 = (1<<30)-1;
const int mod=1000000007;
const int maxn=100005;

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=0;
    while(x>0){
        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,0,sizeof(c));
        memset(d,0,sizeof(d));
        for(int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].h);

        sort(a+1,a+n+1,cmp1);a[1].xx = 1;
        for(int i=2;i<=n;i++){
            if(a[i].x == a[i-1].x) a[i].xx = a[i-1].xx;
            else a[i].xx = i;
        }
        sort(a+1,a+n+1,cmp2);a[1].hh = 1;
        for(int i=2;i<=n;i++){
            if(a[i].h != a[i-1].h) a[i].hh =i;
            else a[i].hh = a[i-1].hh;
        }    
        sort(a+1,a+n+1,cmp3);
        
        //for(int i=1;i<=n;i++)
        //printf("a[%d].x = %d a[%d].h = %d
",i,a[i].xx,i,a[i].hh);
        
        LL ans = 0;
        for(int i=1;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-1)*a[i].xx);
        //    printf("x= %I64d  totalfront=%I64d total=%I64d  ans = %I64d
",x,totalfront,total,ans);
            add(c,a[i].xx,1);
            add(d,a[i].xx,a[i].xx);
        }
        printf("%I64d
",ans);
    }
    return 0;
}