POJ 1990 MooFest

http://poj.org/problem?id=1990

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

Source
USACO 2004 U S Open

题目大意：有n头牛，每头牛有2个值，一个是它的坐标x，另一个是它所能听到的音量的最低要求v，求让所有牛都听到产生的最小音量。音量=（xi-xj)*max(vi,vj).

大概思路：题型：树状数组。
1.如何将题目转换为树状数组可求的模型？这里对每头牛，我们需要求出v小于等于它的牛的数量，然后用n*v。
2.如何求v小于等于它的牛的数量呢？这里一头牛遍历一遍为20000，有20000头牛就需要400000000，所以很明显是不行的。所以需要用到树状数组。
3.为了方便起见，我们先按v排序。我有点不知道怎么说，还是直接解释程序吧。
前面音量和坐标都比它小的：[（x-x1）+（x-x2)+……+(x-xn)]* v[x]= (n *x-Xn) *v[x]
音量比它小但是坐标比它大的：[(x1-x)+(x2-x)+……+(xm-x)]*v[x]=(Xm-m *x) *v[x]
这两个数加起来就是这头牛会产生的音量。
然后具体一点：n怎么求？t1[N]数组用来存储每头牛的n，m怎么求，t2[x]=i-t1[x],因为当前的牛数减去前面的牛就是后面的牛。Xn怎么求？就这么求呗……希望几百年后的自己还能看懂这个程序……

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#define N 20005using namespace std;struct cow{    int v,x;           //v为音量，x为坐标}a[N];long long int c[N],d[N],t1[N],t2[N],s1[N],s2[N];bool cmp(cow a,cow b){    return a.v<b.v;         //按照音量升序排列}void update1(int x,int val){    for(;x<N;x+=(x&(-x)))      //这个函数用来更新音量比它小的牛的数量        c[x]+=val;}long long que1(int x){    long long ret=0;    for(;x;x-=(x&(-x)))        ret+=c[x];    return ret;}void update2(int x,int val)          {    for(;x<N;x+=(x&(-x)))        d[x]+=val;}long long int que2(int x)     //这个是用来求坐标和的。{    long long int ret=0;    for(;x;x-=(x&(-x)))        ret+=d[x];    return ret;}int main(){    int n,i;    while(scanf("%d",&n)!=EOF)    {        for(i=1;i<=n;i++)              //注意树状数组从1开始，0处无意义            scanf("%d%d",&a[i].v,&a[i].x);        sort(a+1,a+n+1,cmp);        memset(c,0,sizeof(c));         //初始化        memset(d,0,sizeof(d));        for(i=1;i<=n;i++)        {            update1(a[i].x,1);            t1[i]=que1(a[i].x);     //这个就是这头牛在它前面的牛的数量            t2[i]=i-t1[i];          //这个是音量比它小但是坐标比它大的牛的数量        }                           //i为所有音量比它小的牛的数量        for(i=1;i<=n;i++)        {            update2(a[i].x,a[i].x);            s1[i]=que2(a[i].x);        //这个是所有x坐标比它小的牛的坐标和            s2[i]=que2(20000)-s1[i];   //这个是所有x坐标比它大的牛的坐标和        }                              //que（20000）为所有出现过的x坐标的总和        long long int ans=0;        for(i=1;i<=n;i++)            ans+=((t1[i]*a[i].x-s1[i])+(s2[i]-t2[i]*a[i].x))*a[i].v;        printf("%lld\n",ans);    }    return 0;}

现在写博客最幸福的时刻就是给我的博文配图了。~(≧▽≦)/~啦啦啦
因为这道题真的不知道怎么说了，呵呵……