POJ1990-MppFest-树状数组

原题链接
MooFest
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 7357 Accepted: 3299
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

题意：奶牛节：N头奶牛每头耳背程度v，坐标x。两牛交流需要音量为distance * max(v(i),v(j))，求所有牛两两交流所需总和？
思路：将奶牛的耳背程度按照升序排列后，从小到大取的话每一次都是已经有的奶牛里最耳背的，就只需要将之前所有的奶牛距离*该奶牛的耳背程度即可，结果就是所有的乘积的加和。比如对于样例来说将最不耳背的奶牛放入后，我们可以通过后面的三个操作得到结果

res += 2 * (|6-5|)res += 3 * (|1-5| + |1-6|)res ++ 4 * (|2-1| + |2-5| + |2-6|)

我们通过观察可以得到一个通式

res += v[i] * (x[i] * 小于等于x[i]的元素个数 - 小于等于x[i]的元素的位置和 - x[i] * 大于x[i]的元素个数 + 大于x[i]的元素的位置和)

我们明显经常用到求区间和的操作，那么我们就可以使用树状数组来快速地实现求区间和的操作。建立一个奶牛个数的BIT数组，和位置和的BIT数组

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 20000 + 10;typedef long long ll;pair<ll,ll> a[maxn];//first代表音量v，second代表位置xvoid add(ll *bit,ll x,ll m){    while(x<maxn){        bit[x] += m;        x += (x & -x);    }}ll sum(ll *bit,ll x){    ll s=0;    while(x>0){        s += bit[x];        x -= (x & -x);    }    return s;}int main(){    ll bita[maxn],bitb[maxn],n;//分别代表位置小于等于a[i].x的元素的个数和这些元素的位置和    ll res = 0;    memset(bita,0,sizeof(bita));    memset(bitb,0,sizeof(bitb));    scanf("%lld",&n);    for(int i=0;i<n;i++) scanf("%lld%lld",&a[i].first,&a[i].second);    sort(a,a+n);    add(bita,a[0].second,1);    add(bitb,a[0].second,a[0].second);    for(int i=1;i<n;i++){        ll lowbitasum=sum(bita,a[i].second);        ll lowbitbsum=sum(bitb,a[i].second);        //res += a[i].first * (a[i].second*lowbitasum - lowbitbsum - a[i].second*(sum(bita,maxn-1)-lowbitasum) + (sum(bitb,maxn-1)-lowbitbsum));        res += a[i].first * (a[i].second * (lowbitasum*2 - sum(bita,maxn-1)) - lowbitbsum*2 + sum(bitb,maxn-1));        add(bita,a[i].second,1);        add(bitb,a[i].second,a[i].second);    }    cout << res << endl;    return 0;}