poj 1990 MooFest（树状数组变形）

MooFest

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 7054 Accepted: 3171

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

题目大意：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；

思路:挂在树状数组专题里。。否则还真的想不到用树状数组做。。。最开始我想的思路是，用结构体存每个点的x，跟v，然后按照v从小到大排序，这样从前面往后枚举的每一个点肯定是当前点最大的，用这个点乘以这个点跟别的点的差的和就好了，关于差的和我一开始想的是直接用现在存在的所有点的坐标和-去当前点x*存在的点的数量。。。点的数量用树状数组维护。。点坐标的和也用树状数组维护。。（我为什么要用树状数组呢？直接一个变量记录点的和不就好了。。。因为后面的点肯定要跟前面点连一次，前面的点不可能跟后面的点连，因为前面的v小。。，所以后面的点肯定跟前面所有点连一次），但是这样直接减是不对的。。。因为距离是两个点相减的绝对值。。用一个变量记录其他现在存在点坐标的和，肯定有比当前点坐标小的，这样一减就是负的了。。。

所以要分情况，先找出在当前点前面的点，以及坐标和，用当前点x*前面点的数量减去前面点的和，就是“绝对值”了。后面的点的数量就是i - 前面的点 - 1（本身），后面点的和就是现在所有点的和（除了当前点）- 减去前面点的和；

和。

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;const int maxn = 1e5;int c[maxn], sum[maxn], n;struct node{    int x, v;    bool operator < (const node &a) const    {        return v < a.v;    }}a[maxn];int lowbit(int k){    return k & -k;}void update1(int pos, int val){    while(pos < maxn)    {        c[pos] += val;        pos += lowbit(pos);    }}void update2(int pos, int val){    while(pos < maxn)    {        sum[pos] += val;        pos += lowbit(pos);    }}int sum1(int pos){    int ans = 0;    while(pos)    {        ans += c[pos];        pos -= lowbit(pos);    }    return ans;}int sum2(int pos){    int ans = 0;    while(pos)    {        ans += sum[pos];        pos -= lowbit(pos);    }    return ans;}int main(){    while(~scanf("%d", &n))    {        memset(c, 0, sizeof(c));        memset(sum, 0, sizeof(sum));        for(int i = 1; i <= n; i++)            scanf("%d%d",&a[i].v, &a[i].x);        sort(a+1, a+1+n);        long long ans = 0, tempsum = 0;        for(int i = 1; i <= n; i++)        {        //  tempsum += a[i].v;            int r = sum1(a[i].x);//记录个数            int t = sum2(a[i].x);//记录坐标和            ans += a[i].v*(r*a[i].x-t+tempsum-t-(i-r-1)*a[i].x);            tempsum += a[i].x;    //这样保证除了当前点            update1(a[i].x, 1);            update2(a[i].x, a[i].x);        }        printf("%lld\n", ans);    }    return 0;}