poj 1990 MooFest（树状数组）

MooFest

Description

Every year, Farmer John’s N (1 <= N <= 20,000) cows attend “MooFest”,a social gathering of cows from around the world. MooFest involves a variety of events including haybale stacking, fence jumping, pin the tail on the farmer, and of course, mooing. When the cows all stand in line for a particular event, they moo so loudly that the roar is practically deafening. After participating in this event year after year, some of the cows have in fact lost a bit of their hearing.

Each cow i has an associated “hearing” threshold v(i) (in the range 1..20,000). If a cow moos to cow i, she must use a volume of at least v(i) times the distance between the two cows in order to be heard by cow i. If two cows i and j wish to converse, they must speak at a volume level equal to the distance between them times max(v(i),v(j)).

Suppose each of the N cows is standing in a straight line (each cow at some unique x coordinate in the range 1..20,000), and every pair of cows is carrying on a conversation using the smallest possible volume.

Compute the sum of all the volumes produced by all N(N-1)/2 pairs of mooing cows.
Input

• Line 1: A single integer, N

• Lines 2..N+1: Two integers: the volume threshold and x coordinate for a cow. Line 2 represents the first cow; line 3 represents the second cow; and so on. No two cows will stand at the same location.
  Output

• Line 1: A single line with a single integer that is the sum of all the volumes of the conversing cows.
  Sample Input

4
3 1
2 5
2 6
4 3
Sample Output

57

题意：一群牛参加完牛的节日后都有了不同程度的耳聋，第i头牛听见别人的讲话，别人的音量必须大于v[i]，当两头牛i，j交流的时候，交流的最小声音为max{v[i],v[j]}*他们之间的距离。现在有n头牛，求他们之间两两交流最少要的音量和。

分析：树状数组。首先将这n头牛按照v值从小到大排序，这样的话，排在后面的牛和排在前面的牛讲话，两两之间所用的音量必定为后面的牛的v值，因此对于每一头牛只需要让它和它前面的所有牛讲话即可。

这样，知道了讲话的v值，还需要知道对于牛i，它前面总共有多少头牛，和在它前面所有牛的位置之和。所以我们需要两个数组，一个a[]记录个数，另一个b[]记录位置之和。

对于每一头牛i讲话的总能量为v[i]*（num*pos[i]-sum）（num为比牛i的v小的牛的个数，sum为比牛i的v小的所有牛的位置之和，具体详见代码）

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define maxn 20010typedef long long LL;struct node{    LL pos,val;} q[maxn];LL a[maxn],b[maxn];//a[]存储个数，b[]存储位置和LL n;LL lowbit(LL x){    return x&-x;}LL update_sum(LL x,LL num)//更新位置之和{    while(x<maxn)    {        b[x]+=num;        x+=lowbit(x);    }}LL get_sum(LL x)//得到位置小于等于x的所有牛的位置之和{    LL sum=0;    while(x)    {        sum+=b[x];        x-=lowbit(x);    }    return sum;}LL update_num(LL x,LL num)//更新个数{    while(x<maxn)    {        a[x]+=num;        x+=lowbit(x);    }}LL get_num(LL x)//得到小于位置x的牛的个数{    LL sum=0;    while(x)    {        sum+=a[x];        x-=lowbit(x);    }    return sum;}bool cmp(node a,node b){    if(a.val==b.val)        return a.pos<b.pos;    return a.val<b.val;}LL absolute(LL x)//求绝对值{    return x<0?-x:x;}int main(){    scanf("%lld",&n);    for(LL i=1; i<=n; ++i)        scanf("%lld%lld",&q[i].val,&q[i].pos);    sort(q+1,q+1+n,cmp);    LL ans=0;    for(LL i=1; i<=n; ++i)    {        ans+=absolute(get_num(q[i].pos-1)*q[i].pos-get_sum(q[i].pos-1))*q[i].val;        //求位置小于q[i]（并且v小于q[i]）的总能量值之和        ans+=absolute((get_num(maxn-1)-get_num(q[i].pos))*q[i].pos-(get_sum(maxn-1)-get_sum(q[i].pos)))*q[i].val;        //求位置大于q[i](并且v小于q[i])的总能量值之和        update_num(q[i].pos,1);//更新        update_sum(q[i].pos,q[i].pos);    }    printf("%lld\n",ans);    return 0;}