POJ 1990 MooFest（树状数组）

MooFest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9006		Accepted: 4073

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】有n头牛，排列成一条直线，给出每头牛在直线上的坐标d。每头牛有一个v，如果牛i和牛j想要沟通的话，它们必须用max(v[i],v[j])，消耗的能量为：max(v[i],v[j]) * 它们之间的距离.

问要使所有的牛之间都能沟通(两两之间)，总共需要消耗多少能量。

【思路】现将v从小到大排列，使得每次取到的是当前最大的v。

c[1]记录当前牛的数量c[2]记录当前所有牛的d之和。（二维树状数组）

题目大意：

给定n头牛的听力和坐标，每两头牛交谈需要（max（v（i），v（j) )*abs(dis[i]-dis[j])),n头牛总共要交谈（n*（n-1）/2）次

思路：

对牛的听力进行从小到大排序，那么对于第i头牛 交谈，需要计算

1：坐标比第i头牛小的牛的数量 a，坐标和b

2：坐标比第i头牛大的牛的数量 c，坐标和d

那么第i头牛交谈总共需要：

(d-c*a[i].y+a*a[i].y-b)*a[i].x

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N = 20000;
struct node
{
    int d, v;
    bool operator <(const node &a)const
        //从小到大排序，使得当前获得的v一定是出现过最大的。
    {
        return v<a.v;
    }
}moo[N + 10];
int c[3][N + 10];
void update(int i, int d, int v)
{
    while (d <= N)
    {
        c[i][d] += v;
        d +=(d&-d);
    }
}
int get_sum(int i, int d)
{
    int res = 0;
    while (d)
    {
        res += c[i][d];
        d -= (d&-d);
    }
    return res;
}
int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        memset(c, 0, sizeof(c));
        for (int i = 1; i <= n; i++)
            scanf("%d%d", &moo[i].v, &moo[i].d);
        sort(moo + 1, moo + 1 + n);
        int sum = 0;//记录所有坐标之和
        long long int ans = 0;
        for (int i = 1; i <= n; i++)
        {
            int d = moo[i].d;
            sum += d;
            update(1, d, 1);//c[1]记录牛数量
            update(2, d, d);//c[2]记录牛坐标之和
            int n1 = get_sum(1, d);//在i牛及他前面有多少头
            int n2 = get_sum(2, d);//在i牛及他前面的牛坐标和为多少
            int tmp1 = n1 * d - n2;//i左边的坐标差
            int tmp2 = sum - n2 - d * (i - n1);//i右边的坐标差
            ans += (long long int)(tmp1 + tmp2)*moo[i].v;
            //不用longlong会溢出
        }
        printf("%lld
", ans);
    }
    return 0;
}