SOJ-2728（树状数组）

/******************************************************************************************************
 ** Copyright (C) 2011.07.01-2013.07.01
 ** Author: famousDT <[email protected]>
 ** Edit date: 2011-08-06
******************************************************************************************************/
#include <stdio.h>
#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll
#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10
#include <vector>
#include <queue>
#include <map>
#include <time.h>
#include <set>
#include <stack>
#include <string>
#include <iostream>
#include <assert.h>
#include <string.h>//memcpy(to,from,count
#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprll
#include <algorithm>
using namespace std;

typedef long long ll;

#define PI acos(-1)
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))
#define FABS(a) ((a) >= 0 ? (a) : (-(a)))
#define MAX_INT 0xfffffff

/***********************************解题报告*******************************************
 ** 树状数组
 ** 题目描述：很多牛站在x轴上，牛们要聊天，需要一定的v值；两个牛a，b通话时，
    花费=距离 * max(v(a), v(b)），求所有牛的花费和。
 ** 思路：
    按照v值从小到大排个序，对于第i头牛：
    count[i]记录在它前面且x坐标比它小的牛的头数之和
    total[i]记录在它前面且x坐标比它小的牛的x坐标之和
    alltotal记录在它前面总的牛的x坐标之和
    结果可以表示为：
    v * (count[i] * x - total[i] + alltotal - total[i] - (i - count[i] - 1) * x)
    其中count[i] * x - total[i]表示x坐标比它小的距离之和，
    alltotal - total[i] - (i - count[i] - 1) * x表示x坐标比它大的距离之和
    count[i]用c[0][]存储计算，total[i]用c[1][]存储计算
***************************************************************************************/

#define INDEX 20001

ll number;
ll c[2][INDEX];

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

void add(ll index, ll i, ll val)
{
    while (i <= number) {
        c[index][i] += val;
        i += lowbit(i);
    }
}

ll sum(ll index, ll i)
{
    ll s = 0;
    while (i > 0) {
        s += c[index][i];
        i -= lowbit(i);
    }
    return s;
}

struct cows
{
    ll v;
    ll x;
} cow[INDEX];

bool cmp(cows a, cows b)
{
    return a.v < b.v;
}

int main()
{
    int i, n;
    while (cin>>n) {
        ll max = 0;
        for (i = 1; i <= n; ++i) {
            cin>>cow[i].v>>cow[i].x;
            if (cow[i].x > max) max = cow[i].x;
        }
        number = max;
        for (i = 1; i <= max; ++i)
            c[0][i] = c[1][i] = 0;
        sort(cow + 1, cow + n + 1, cmp);
        ll ans = 0;
        ll alltotal = 0;
        for (i = 1; i <= n; ++i) {
            ll pre = sum(0, cow[i].x);
            ll all = sum(1, cow[i].x);
            ans += cow[i].v * (pre * cow[i].x - all + alltotal - all - (i - pre - 1) * cow[i].x);
            alltotal += cow[i].x;
            add(0, cow[i].x, 1);
            add(1, cow[i].x, cow[i].x);
        }
        printf("%lld\n", ans);
    }
    return 0;
}