114. Telecasting station
time limit per test: 0.25 sec.
memory limit per test: 4096 KB
Every city in Berland is situated on Ox axis. The government of the country decided to build new telecasting station. After many experiments Berland scientists came to a conclusion that in any city citizensdispleasure is equal to product of citizens amount in it by distance between city and TV-station. Find such point on Ox axis for station so that sum of displeasures of all cities is minimal.
Input
Input begins from line with integer positive number N (0<N<15000) – amount of cities in Berland. Following N pairs (X, P) describes cities (0<X, P<50000), where X is a coordinate of city and P is an amount of citizens. All numbers separated by whitespace(s).
Output
Write the best position for TV-station with accuracy 10-5.
Sample Input
4 1 3 2 1 5 2 6 2
Sample Output
3.00000
把权值看成有多少个人,求中位数。
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <algorithm> 5 6 using namespace std; 7 8 #define maxn 15005 9 10 struct node { 11 12 int x,p; 13 }; 14 15 int n; 16 node s[maxn]; 17 18 bool cmp(node a,node b) { 19 return a.x < b.x; 20 } 21 22 int main() { 23 //freopen("sw.in","r",stdin); 24 25 scanf("%d",&n); 26 27 int sum = 0; 28 for(int i = 1; i <= n; ++i) { 29 scanf("%d%d",&s[i].x,&s[i].p); 30 sum += s[i].p; 31 } 32 33 sort(s + 1,s + n + 1,cmp); 34 35 int now = 0; 36 double m1,m2; 37 for(int i = 1; i <= n; ++i) { 38 now += s[i].p; 39 if(now >= sum / 2) { 40 m1 = s[i].x; 41 if(now >= sum / 2 + 1) m2 = s[i].x; 42 else m2 = s[i + 1].x; 43 break; 44 } 45 46 } 47 48 if(sum % 2) printf("%.5f ",m2); 49 else printf("%.5f ",(m1 + m2) / 2); 50 51 return 0; 52 53 54 55 56 }