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.
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.
* 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
我觉得我需要收回当初说树状数组比线段树简单这句话。。太坑了，一题比一题坑。。完全按题解写的
题意：给v【i】，x【i】要求所以的牛的音量和即x【i】-x【j】*max（v【i】，v【j】）之和
题解：两个树状数组数组一起使用，一个求x之前的比x坐标小的数(a)，一个求x之前的比x坐标小的坐标和(b);
那么比x小的坐标和x的坐标的总坐标差是a*(e[i].x)-b;比x大的坐标和x的坐标的总坐标差是总坐标-b-(i-1-a)*e[i].x
#include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<iomanip> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define pi acos(-1) #define ll long long #define mod 1000000007 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 using namespace std; const double g=10.0,eps=1e-9; const int N=20000+5,maxn=32000+5,inf=0x3f3f3f3f; int s[2][N]; struct edge{ ll v,x; }e[N]; bool comp(const edge &a,const edge &b) { return a.v<b.v; } void add(int i,ll x,int d) { while(i<=N){ s[d][i]+=x; i+=i&(-i); } } ll sum(int i,int d) { ll ans=0; while(i>0){ ans+=s[d][i]; i-=i&(-i); } return ans; } int main() { ios::sync_with_stdio(false); cin.tie(0); // cout<<setiosflags(ios::fixed)<<setprecision(2); int n; while(cin>>n){ memset(s,0,sizeof s); for(int i=1;i<=n;i++)cin>>e[i].v>>e[i].x; sort(e+1,e+1+n,comp); ll ans=0; for(int i=1;i<=n;i++) { ll a=sum(e[i].x,0),b=sum(e[i].x,1); // cout<<sum(N,1)-b-(i-1-a)*e[i].x<<endl; ans+=(a*e[i].x-b+sum(N,1)-b-(i-1-a)*e[i].x)*e[i].v; add(e[i].x,1,0);//0是比x小的牛的个数 add(e[i].x,e[i].x,1);//1是比x小的牛的距离和 } cout<<ans<<endl; } return 0; }