MooFest
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 9006 | Accepted: 4073 |
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】有n头牛，排列成一条直线，给出每头牛在直线上的坐标d。每头牛有一个v，如果牛i和牛j想要沟通的话，它们必须用max(v[i],v[j])，消耗的能量为：max(v[i],v[j]) * 它们之间的距离.
问要使所有的牛之间都能沟通(两两之间)，总共需要消耗多少能量。
【思路】现将v从小到大排列，使得每次取到的是当前最大的v。
c[1]记录当前牛的数量c[2]记录当前所有牛的d之和。（二维树状数组）
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
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<algorithm> 5 using namespace std; 6 const int N = 20000; 7 struct node 8 { 9 int d, v; 10 bool operator <(const node &a)const 11 //从小到大排序，使得当前获得的v一定是出现过最大的。 12 { 13 return v<a.v; 14 } 15 }moo[N + 10]; 16 int c[3][N + 10]; 17 void update(int i, int d, int v) 18 { 19 while (d <= N) 20 { 21 c[i][d] += v; 22 d +=(d&-d); 23 } 24 } 25 int get_sum(int i, int d) 26 { 27 int res = 0; 28 while (d) 29 { 30 res += c[i][d]; 31 d -= (d&-d); 32 } 33 return res; 34 } 35 int main() 36 { 37 int n; 38 while (~scanf("%d", &n)) 39 { 40 memset(c, 0, sizeof(c)); 41 for (int i = 1; i <= n; i++) 42 scanf("%d%d", &moo[i].v, &moo[i].d); 43 sort(moo + 1, moo + 1 + n); 44 int sum = 0;//记录所有坐标之和 45 long long int ans = 0; 46 for (int i = 1; i <= n; i++) 47 { 48 int d = moo[i].d; 49 sum += d; 50 update(1, d, 1);//c[1]记录牛数量 51 update(2, d, d);//c[2]记录牛坐标之和 52 int n1 = get_sum(1, d);//在i牛及他前面有多少头 53 int n2 = get_sum(2, d);//在i牛及他前面的牛坐标和为多少 54 int tmp1 = n1 * d - n2;//i左边的坐标差 55 int tmp2 = sum - n2 - d * (i - n1);//i右边的坐标差 56 ans += (long long int)(tmp1 + tmp2)*moo[i].v; 57 //不用longlong会溢出 58 } 59 printf("%lld ", ans); 60 } 61 return 0; 62 }