SGU114-Telecasting station

114. Telecasting station time limit per test: 0.5 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 citizens displeasure 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

题意是说，有个国家的城市之间要建个公交（我英语不好、翻译可能会跑偏、也可能是电视塔、也可能是别的、我也不知道是啥），然后有一个不满意度，咱们自己想想也知道，肯定都喜欢把地铁站设在自己家门口。然后这个不满意度=城市的人数*这个站的位置。。。然后的然后、我也不会做这个题，直接上理论：数学家说，这是一个带权中位数问题，具体内容见 http://baike.baidu.com/link?url=NZbFTsJCCn_jsx8W24aLu6XGEVdpDH0hpO_SIzeCF7vFi_Nz-xV4pe7tzuJBpQsQSzoa719FmHYb3lr7QKV3q_   这个讲的很明白了。然后，下面的代码就自然手到擒来了！！！   #include<iostream> #include<string.h> #include<stdio.h> #include<ctype.h> #include<algorithm> #include<stack> #include<queue> #include<set> #include<math.h> #include<vector> #include<map> #include<deque> #include<list> using namespace std; struct N { double x; int p; } a[15009]; int b[15009]; int cmp(N a,N b) { return a.x<b.x; } int main() { int n; while(scanf("%d",&n)!=EOF) { memset(b,0,sizeof(b)); for(int i=0; i<n; i++) scanf("%d%d",&a[i].x,&a[i].p); sort(a,a+n,cmp); int s=0; for(int i=0; i<n; i++) { s+=a[i].p; b[i]=s; } int mid=s/2,w; for(int i=0; i<n; i++) { if(b[i]>=mid) { w=i; break; } } printf("%d\n",a[w].x); } return 0; }