POJ 1990 树状数组

弱第一次知道了树状数组与线段树的差别。。。。实在太弱了。

言归正传：题目：

MooFest

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 7683 Accepted: 3455

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

43 12 52 64 3

Sample Output

57

首先，关于树状数组，见这篇博客：http://blog.csdn.net/int64ago/article/details/7429868

讲的很详细。。。本弱就不自己解释了。

此题需反复多次进行查找，适合使用树状数组

先对所有牛按照位置进行排序，然后按照牛声音的高低建两棵树，一棵存储牛的只数，另一棵存放距离。按照位置进行排序后，保证每次处理都能分清楚牛的前后关系，以方便计算距离。

代码：

//By Sean Chen#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#define MAX 20005using namespace std;int lowbit(int x)         //树状数组的重要操作{    return x&(-x);}struct data{    int x,w;}cow[MAX];int bitNum[MAX],bitDis[MAX],n;bool cmp(data a,data b){    return a.x<b.x;}void add(int i,int *ar,int w)             //添加结点{    while(i<MAX)    {        ar[i]+=w;        i+=lowbit(i);    }    return;}long long sum(int i,int *ar){    long long ans=0;    while(i>0)    {        ans+=ar[i];        i-=lowbit(i);    }    return ans;}int main(){    int i;    long long totcow,totdis;    scanf("%d",&n);    for(i=0;i<n;i++)        scanf("%d%d",&cow[i].w,&cow[i].x);    sort(cow,cow+n,cmp);    long long ans=0;    memset(bitNum,0,sizeof(bitNum));    memset(bitDis,0,sizeof(bitDis));    for(i=0;i<n;i++)                     //向左计算    {        totcow=sum(cow[i].w-1,bitNum);        totdis=sum(cow[i].w-1,bitDis);        ans+=(totcow*cow[i].x-totdis)*cow[i].w;        add(cow[i].w,bitNum,1);        add(cow[i].w,bitDis,cow[i].x);    }    memset(bitNum,0,sizeof(bitNum));    memset(bitDis,0,sizeof(bitDis));    for(i=n-1;i>=0;i--)         //向右计算    {        totcow=sum(cow[i].w,bitNum);        totdis=sum(cow[i].w,bitDis);        ans+=(totdis-totcow*cow[i].x)*cow[i].w;        add(cow[i].w,bitNum,1);        add(cow[i].w,bitDis,cow[i].x);    }    printf("%lld\n",ans);    return 0;}