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
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #define int long long 7 #define maxn 50+10 8 #define rep(i,s,e) for(register int i=s;i<=e;++i) 9 #define dwn(i,s,e) for(register int i=s;i>=e;--i) 10 using namespace std; 11 inline int read() 12 { 13 int x=0,f=1; 14 char c=getchar(); 15 while(c<'0'||c>'9') {if(c=='-') f=-1; c=getchar();} 16 while(c>='0'&&c<='9') {x=x*10+c-'0'; c=getchar();} 17 return f*x; 18 } 19 inline void write(int x) 20 { 21 if(x<0) {putchar('-'); x=-x;} 22 if(x>9) write(x/10); 23 putchar(x%10+'0'); 24 } 25 int n; 26 int dis[maxn][maxn]; 27 struct node 28 { 29 int x,y; 30 }a[maxn]; 31 signed main() 32 { 33 n=read(); 34 rep(i,1,n) 35 { 36 a[i].x=read();a[i].y=read(); 37 rep(j,1,i-1) dis[i][j]=dis[j][i]=(abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)+1)/2; 38 } 39 rep(k,1,n) 40 rep(i,1,n) 41 rep(j,1,n) 42 if(dis[i][j]>max(dis[i][k],dis[k][j])) 43 dis[i][j]=dis[j][i]=max(dis[i][k],dis[k][j]); 44 int ans=0; 45 rep(i,1,n) 46 rep(j,1,n) 47 ans=max(ans,dis[i][j]); 48 write(ans); 49 return 0; 50 }
请各位大佬斧正（反正我不认识斧正是什么意思）