Computer
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30767 Accepted Submission(s):
3802
Problem Description
A school bought the first computer some time ago(so
this computer's id is 1). During the recent years the school bought N-1 new
computers. Each new computer was connected to one of settled earlier. Managers
of school are anxious about slow functioning of the net and want to know the
maximum distance Si for which i-th computer needs to send signal (i.e. length of
cable to the most distant computer). You need to provide this information.
Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4.
Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4.
Input
Input file contains multiple test cases.In each case
there is natural number N (N<=10000) in the first line, followed by (N-1)
lines with descriptions of computers. i-th line contains two natural numbers -
number of computer, to which i-th computer is connected and length of cable used
for connection. Total length of cable does not exceed 10^9. Numbers in lines of
input are separated by a space.
Output
For each case output N lines. i-th line must contain
number Si for i-th computer (1<=i<=N).
Sample Input
5
1 1
2 1
3 1
1 1
Sample Output
3
2
3
4
4
Author
scnu
题目大意 求树上每个点到树上另一个点的最长距离
题解
树形dp
设一个点u到另一个点v,为u出发的最长距离。那么v要么在u的子树里,要么不在子树里...(废话!)
dp[i][0]表示i到它子树的最深长度。dp[i][1]表示i到其子树的次深长度。dp[i][2]表示从i的爸爸出发到达的最深长度。
那么dp[u][0]=max(dp[v][0]+edge[i].z)要推导出一个点到它子树的最深长度,必须要知道u这个点的儿子到他们子树
的最深深度,所以dfs要倒着从根节点递推。而dp[i][2]是怎样推的呢。dp[v][2]=max(dp[u][2],dp[u][0]/dp[u][1])+edge[i].z;
(u是v的爸爸)当v这个点是u到达子树最深节点路径上一个点时,那么dp[v][2]=max(dp[u][2],dp[u][1])+z,否则就是
另一种情况。
代码
#include<iostream> #include<cstring> #include<cstdio> #define maxn 10005 using namespace std; int n,y,w,sumedge; int head[maxn],dp[maxn][3]; struct Edge{ int x,y,z,nxt; Edge(int x=0,int y=0,int z=0,int nxt=0): x(x),y(y),z(z),nxt(nxt){} }edge[maxn]; void add(int x,int y,int z){ edge[++sumedge]=Edge(x,y,z,head[x]); head[x]=sumedge; } void dfs1(int x){ for(int i=head[x];i;i=edge[i].nxt){ int v=edge[i].y; dfs1(v); int gg=dp[v][0]+edge[i].z; if(gg>=dp[x][0]){ dp[x][1]=dp[x][0]; dp[x][0]=gg; }else if(gg>=dp[x][1])dp[x][1]=gg; } } void dfs2(int x){ for(int i=head[x];i;i=edge[i].nxt){ int v=edge[i].y; if(dp[x][0]==dp[v][0]+edge[i].z) dp[v][2]=max(dp[x][1],dp[x][2])+edge[i].z; else dp[v][2]=max(dp[x][2],dp[x][0])+edge[i].z; dfs2(v); } } int main(){ while(~scanf("%d",&n)){ memset(head,0,sizeof(head)); memset(dp,0,sizeof(dp)); sumedge=0; for(int i=2;i<=n;i++){ scanf("%d%d",&y,&w); add(y,i,w); } dfs1(1); dfs2(1); for(int i=1;i<=n;i++) printf("%d ",max(dp[i][0],dp[i][2])); } return 0; }