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POJ 1990——MooFest(2个树状数组)
MooFest
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 5235 | Accepted: 2260 |
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
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题目大意:
给定n头牛的听力和坐标,每两头牛交谈需要(max(v(i),v(j) )*abs(dis[i]-dis[j])),n头牛总共要交谈(n*(n-1)/2)次
思路:
对牛的听力进行从小到大排序,那么对于第i头牛 交谈,需要计算
1:坐标比第i头牛小的牛的数量 a,坐标和b
2:坐标比第i头牛大的牛的数量 c,坐标和d
那么第i头牛交谈总共需要:
(d-c*a[i].y+a*a[i].y-b)*a[i].x
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #define M 20000+10 #define ll __int64 using namespace std; ll b[M],c[M]; struct node { int x,y; bool operator<(const node&a)const { return x<a.x; } }a[M]; void update(int x,int v,ll *h) { for(int i=x;i<=M;i+=i&-i){ h[i]+=v; } } ll getsum(int x,ll *h) { ll sum=0; for(int i=x;i>0;i-=i&-i){ sum+=h[i]; } return sum; } int main() { int n; scanf("%d",&n); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); for(int i=1;i<=n;++i){ scanf("%d %d",&a[i].x,&a[i].y); } sort(a+1,a+1+n); ll ans=0; for(int i=1;i<=n;++i){ ll x1=(ll)(getsum(a[i].y,b)*a[i].y-getsum(a[i].y,c)); ll x2=(ll)(getsum(M,c)-getsum(a[i].y,c)-(i-1-getsum(a[i].y,b))*a[i].y); ans+=(ll)(x1+x2)*a[i].x; update(a[i].y,1,b); update(a[i].y,a[i].y,c); } printf("%I64d\n",ans); return 0; }
POJ 1990——MooFest(2个树状数组)
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