Educational Codeforces Round 15 C. Cellular Network(二分)
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题目链接:C. Cellular Network
让每一个城市都可以被供给,也就是说每一个发电站的距离向左右延伸后要把城市最小~最大范围全部覆盖到,想了一会儿感觉是找每一个城市的最近供给站的距离,然后取最大值作为标准,然后进行二分……,然后怎么找这个最近的供给站呢?还是二分……,找的时候注意边界就可以了
代码:
#include<bits/stdc++.h>using namespace std;#define INF 0x3f3f3f3f#define MM(x,y) memset(x,y,sizeof(x))#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)typedef pair<int,int> pii;typedef long long LL;const double PI=acos(-1.0);const int N=100010;int n,m;int c[N];int d[N];int vis[N];LL min_dis[N];inline LL ABS(const LL &a){return a>0?a:-a;}int main(void){int i,j,k;while (~scanf("%d%d",&n,&m)){for (i=0; i<n; ++i){scanf("%d",&c[i]);}for (i=0; i<m; ++i){scanf("%d",&d[i]);}LL dx=0;for (i=0; i<n; ++i){int l=lower_bound(d,d+m,c[i])-d;int r=upper_bound(d,d+m,c[i])-d-1;if(l>=m)l=m-1;if(r>=m)r=m-1;if(r<0)r=0;min_dis[i]=min(ABS((LL)c[i]-d[l]),ABS((LL)c[i]-d[r]));if(min_dis[i]>dx)dx=min_dis[i];}int nm=n+m;LL l=0,r=2LL*1e9,mid,ans;while (l<=r){mid=(l+r)>>1;if(mid>=dx){r=mid-1;ans=mid;}elsel=mid+1;}printf("%I64d\n",ans);}return 0;}
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