LOJ P10015 扩散 题解

每日一题 day49 打卡

Analysis

用dis数组记录每两个点之间的时间，再用一个传递闭包来维护最小的时间就好了

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #define int long long
 #define maxn 50+10
 #define rep(i,s,e) for(register int i=s;i<=e;++i)
 #define dwn(i,s,e) for(register int i=s;i>=e;--i)
 using namespace std;
 inline int read()
 {
     int x=,f=;
     char c=getchar();
     while(c<''||c>'') {if(c=='-') f=-; c=getchar();}
     while(c>=''&&c<='') {x=x*+c-''; c=getchar();}
     return f*x;
 }
 inline void write(int x)
 {
     if(x<) {putchar('-'); x=-x;}
     if(x>) write(x/);
     putchar(x%+'');
 }
 int n;
 int dis[maxn][maxn];
 struct node
 {
     int x,y;
 }a[maxn];
 signed main()
 {
     n=read();
     rep(i,,n)
     {
         a[i].x=read();a[i].y=read();
         rep(j,,i-) dis[i][j]=dis[j][i]=(abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)+)/;
     }
     rep(k,,n)
         rep(i,,n)
             rep(j,,n)
                 if(dis[i][j]>max(dis[i][k],dis[k][j]))
                     dis[i][j]=dis[j][i]=max(dis[i][k],dis[k][j]);
     int ans=;
     rep(i,,n)
         rep(j,,n)
             ans=max(ans,dis[i][j]);
     write(ans);
     return ;
 }

请各位大佬斧正（反正我不认识斧正是什么意思）