Educational Codeforces Round 15 Cellular Network
Cellular Network
题意:
给n个城市,m个加油站,要让m个加油站都覆盖n个城市,求最小的加油范围r是多少。
题解:
枚举每个城市,二分查找最近的加油站,每次更新答案即可,注意二分的时候不要越界oil数组,上下界都不要越。还有,int坑死人,以后绝对全用long long!!!
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const ll LINF=0x3f3f3f3f3f3f3f3f;
#define PU puts("");
#define PI(A) cout<<A<<endl
#define SI(N) cin>>N
#define SII(N,M) cin>>N>>M
#define cle(a,val) memset(a,(val),sizeof(a))
#define rep(i,b) for(int i=0;i<(b);i++)
#define Rep(i,a,b) for(int i=(a);i<=(b);i++)
#define reRep(i,a,b) for(int i=(a);i>=(b);i--)
#define dbg(x) cout <<#x<<" = "<<x<<endl
#define PIar(a,n) rep(i,n)cout<<a[i]<<" ";cout<<endl;
#define PIarr(a,n,m) rep(aa,n){rep(bb, m)cout<<a[aa][bb]<<" ";cout<<endl;}
const double EPS= 1e-9 ;
/* ///////////////////////// C o d i n g S p a c e ///////////////////////// */
const int MAXN= 100000 + 9 ;
ll city[MAXN],oil[MAXN];
ll N,M;
int main()
{
iostream::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
while(SII(N,M))
{
rep(i,N)SI(city[i]);
rep(i,M)SI(oil[i]);
ll ans=0;
rep(i,N)
{
ll t=lower_bound(oil,oil+M,city[i])-oil;
ll d=LINF;
if (t<M)
d=abs(oil[t]-city[i]);
ll d2=LINF;
if (t-1>=0)
d2=abs(oil[t-1]-city[i]);
ans=max(ans,min(d,d2));
}
PI(ans);
}
return 0;
}