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POJ 1990 MooFest(树状数组)
MooFest
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 5395 | Accepted: 2329 |
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.
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
先按听力值从小到大排个序,保证当前的牛与前面的联系时,取当前牛的听力值,再用两个树状数组,一个记录所有所有坐标点的位置个数,这样就可以查询大于当前坐标的个数和小于当前坐标的个数,另一个记录所有坐标点的和,这样就可以查询小于和大于当前坐标的所有坐标的和,最后用当前听力值*(大于当前坐标的坐标和—大于当前坐标的个数*当前坐标+小于于当前坐标的个数*当前坐标—小于当前坐标的坐标和)。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn=20000+1000; struct node { int x; int y; }a[maxn]; long long b[maxn]; int c[maxn]; bool cmp(node p,node q) { return p.x<q.x; } int low(int k) { return (k&(-k)); } void update(int k,int v) { while(k<maxn) { b[k]+=v; k+=low(k); } } void update2(int k,int v) { while(k<maxn) { c[k]+=v; k+=low(k); } } long long getsum(int k) { long long ans=0; while(k>0) { ans+=b[k]; k-=low(k); } return ans; } int getsum2(int k) { int ans=0; while(k>0) { ans+=c[k]; k-=low(k); } return ans; } int main() { long long ans; int n; while(~scanf("%d",&n)) { memset(a,0,sizeof(a)); for(int i=0;i<n;i++) { scanf("%d%d",&a[i].x,&a[i].y); } ans=0; sort(a,a+n,cmp); long long temp=0; for(int i=0;i<n;i++) { long long x1=getsum(a[i].y);//小于当前坐标的坐标和 long long y1=temp-x1;//大于当前坐标的坐标和 int x2=getsum2(a[i].y);//小于当前坐标的个数 int y2=i-x2;//大于当前坐标的个数 ans+=(long long)a[i].x*(y1-a[i].y*y2); ans+=(long long)a[i].x*(a[i].y*x2-x1); update(a[i].y,a[i].y); update2(a[i].y,1); temp+=a[i].y; } printf("%I64d\n",ans); } return 0; }
POJ 1990 MooFest(树状数组)
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