hdu 2196 computer (树形dp）

分类： acm 动态规划 2013-07-21 15:53 93人阅读评论(0) 收藏举报
Computer
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1885 Accepted Submission(s): 934

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.

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

Recommend
lcy

解题思路：
题意是求树中每个点到所有叶子节点的距离的最大值是多少。
由于对于一个节点来说，可能得到的距离最大的值的路径来自他的子树，或者从他的父节点过来，所以用两次DFS。
第一次DFS求出所有节点在他的子树范围内到叶子节点距离的最大值和第二大的值，第二次DFS更新从父节点过来的情况就可以了。
因为如果只存最大值的话，判断一个点的从父节点过来的最大值，那么如果他的父节点存的最大值正好是从该点过来的，那么就失去了从父节点过来的状态，所以要记录最大的两个值。

#include <cstdio>
#include <algorithm>
using namespace std;
const long M=21000;
long dp[M][2],id[M][2],n;
long head[M],Next[M],to[M],val[M],cnt;
void init_edge(){
	for (long i=1;i<=n;++i) head[i]=-1;
	cnt=0;
}
void add_edge(long u,long v,long c){
	Next[++cnt]=head[u]; head[u]=cnt; to[cnt]=v;
	val[cnt]=c;
}

void Swap(long x){
	if (dp[x][1]>=dp[x][0]){
		swap(dp[x][1],dp[x][0]);
		swap(id[x][1],id[x][0]);
	}
}

void dfs1(long u,long f){
	dp[u][0]=0;  dp[u][1]=0;
	for (long i=head[u];i!=-1;i=Next[i]){
		long v=to[i];
		if (v==f) continue;
		dfs1(v,u);
		if (dp[v][0]+val[i]>dp[u][1]){
			dp[u][1]=dp[v][0]+val[i];
			id[u][1]=v;
			Swap(u);
		}
	}
}

void dfs2(long u,long f){
	for (long i=head[u];i!=-1;i=Next[i]){
		long v=to[i];
		if (v==f) continue;
		if (v==id[u][0]){
			if (dp[u][1]+val[i]>dp[v][1]){
				dp[v][1]=dp[u][1]+val[i];
				id[v][1]=u;
				Swap(v);
			}
		}
		else{
			if (dp[u][0]+val[i]>dp[v][1]){
				dp[v][1]=dp[u][0]+val[i];
				id[v][1]=u;
				Swap(v);
			}
		}
		dfs2(v,u);
	}
}

int main(){
	while (~scanf("%d",&n)){
		init_edge();
		for (long i=2;i<=n;++i){
			long u,c;
			scanf("%d%d",&u,&c);
			add_edge(u,i,c); add_edge(i,u,c);
		}
		dfs1(1,-1);

		dfs2(1,-1);
		for (long i=1;i<=n;++i)
			printf("%d\n",dp[i][0]);
	}
	return 0;
}

补充：软件开发 , C++ ,