POJ 1990 MooFest (树状数组)
MooFest
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 8063 | Accepted: 3640 |
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
题意: 给n 头牛, 每头牛站成一排, 每个牛都有一个 听力值, 现在 如果两头牛想要相互说话, 则 必须满足 听力值达到 max(v[i],v[j])才可以;
问使得梅梅都可以相互交流 最少的花费是多少;
思路: 使用树状数组 使查询控制在n*log(n) 内 建立两个 树状数组, 分别是距离和 听力值的; 按照 听力值上升排序, 这样 每次 的都是最大的;
要查询 当前为值 左边的 距离比他小的个数, 和右边比他的个数;
则 左边就是 x*a-b 右边是 b-a*x; 要用long long 数据大
#include <iostream> #include <stdio.h> #include <cmath> #include <queue> #include <cstring> #include <algorithm> typedef long long ll; using namespace std; const int N =20019; ll tree_n[N]; ll tree_x[N]; ll n; struct node{ ll v, x; }a[N]; int cmp(node a,node b) { return a.v<b.v; } void add(ll tree[],int k,ll num)//维护添加值 { while(k<=N) { tree[k]+=num; k+=k&(-k);// lowbit(k)= k&(-k); } } ll Sum(int x,ll tree[])//求区间[1,x] 的和 { ll sum=0; while(x>0) { sum+=tree[x]; x-=x&(-x); } return sum; } ll Get_sum(ll tree[],int x,int y)// x-y区间的和 { return Sum(y-1,tree)-Sum(x-1,tree); } int main() { // freopen("input.txt","r",stdin); while(cin>>n) { memset(tree_n,0,sizeof(tree_n)); memset(tree_x,0,sizeof(tree_x)); memset(a,0,sizeof(a)); for(int i=1;i<=n;i++) scanf("%lld %lld",&a[i].v,&a[i].x); sort(a+1,a+n+1,cmp); ll ans=0; for(int i=1;i<=n;i++) { ll left=a[i].x*Sum(a[i].x-1,tree_n)-Sum(a[i].x-1,tree_x); ll right=Get_sum(tree_x,a[i].x,N)-a[i].x*Get_sum(tree_n,a[i].x,N); ans+= (left+right) * a[i].v; add(tree_n,a[i].x,1);// v值 add(tree_x,a[i].x,a[i].x);// 距离 } cout<<ans<<endl; } return 0; }
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