[POI2008]Sta
Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 1889 Solved: 729
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Description
给出一个N个点的树,找出一个点来,以这个点为根的树时,所有点的深度之和最大
Input
给出一个数字N,代表有N个点.N<=1000000 下面N-1条边.
Output
输出你所找到的点,如果具有多个解,请输出编号最小的那个.
Sample Input
8
1 4
5 6
4 5
6 7
6 8
2 4
3 4
1 4
5 6
4 5
6 7
6 8
2 4
3 4
Sample Output
7
HINT
题解：这是一道裸题
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cmath> 7 #include<vector> 8 9 #define N 1000007 10 #define ll long long 11 using namespace std; 12 inline int read() 13 { 14 int x=0,f=1;char ch=getchar(); 15 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 16 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 17 return x*f; 18 } 19 20 int n,m,id; 21 int cnt,hed[N],rea[N<<1],nxt[N<<1]; 22 ll sum[N],dep[N],siz[N]; 23 24 void add(int u,int v) 25 { 26 nxt[++cnt]=hed[u]; 27 hed[u]=cnt; 28 rea[cnt]=v; 29 } 30 void add_two_way(int x,int y) 31 { 32 add(x,y); 33 add(y,x); 34 } 35 void dfs1(int u,int fa) 36 { 37 siz[u]=1; 38 for (int i=hed[u];i!=-1;i=nxt[i]) 39 { 40 int v=rea[i]; 41 if (v==fa) continue; 42 dep[v]=dep[u]+1; 43 dfs1(v,u); 44 sum[u]+=sum[v],siz[u]+=siz[v]; 45 } 46 sum[u]+=dep[u]; 47 } 48 void dfs2(int u,int fa) 49 { 50 for (int i=hed[u];~i;i=nxt[i]) 51 { 52 int v=rea[i]; 53 if (v==fa) continue; 54 sum[v]=sum[u]-siz[v]+n-siz[v]; 55 dfs2(v,u); 56 } 57 } 58 int main() 59 { 60 memset(hed,-1,sizeof(hed)); 61 n=read(); 62 for (int i=1;i<n;i++) 63 add_two_way(read(),read()); 64 dfs1(1,0),dfs2(1,0); 65 for (int i=1;i<=n;i++) if (sum[i]>sum[id]) id=i; 66 printf("%d ",id); 67 }