dig accept log names problem other vector http tchar

Computer

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7324 Accepted Submission(s): 3627

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 題目大意：給定N個點的無根樹，求每個點到它最遠點的距離。 試題分析：設dp[i]表示i到最遠點的距離，我們發現dp[i]=max(dp[i->son]+1,dp[fa[i]]+1) 　　　　　但是這個轉移方程在父節點的最長路如果是走到i的子樹的那麽就出錯了。 　　　　　於是我們添加一維狀態:dp[i][2] 表示i號節點的子樹中最遠點與不在子樹中最遠點。 　　　　　dp[i][0]=max(dp[i->son][0]+1) 　　　　　dp[i][1]=max(dp[fa[i]][1]+1,dp[fa[i]][0]+1) 　　　　　列出轉移方程，發現我們的顧慮並沒有消除，我們不確定父節點的在子樹中最大的是不是i的子樹。 　　　　　為了解決這個問題，我們將狀態變成dp[i][3]表示i到子樹中最遠點，i到子樹中次遠點，不在i的子樹中與i距離最遠的點 　　　　　那麽dp[i][0]=max(dp[i->son][0]+1) 　　　　　　　dp[i][1]=max(dp[i->son][0]+1) (i->son不滿足max(dp[i->son][0])) 　　　　　　　dp[i][2]=max(dp[fa[i]][2]+1,dp[fa[i]][0]+1) (dp[fa[i]][0]的節點不在i的子樹中) 　　　　　　　　　　 max(dp[fa[i]][2]+1,dp[fa[i]][1]+1) (如果在就要退而求其次了)
代碼：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c==‘-‘) f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-‘0‘;
	return x*f;
}
const int MAXN=20001;
const int INF=999999;
int N,M; int len[MAXN];
vector<int> vec[MAXN];
struct data{
    int Son,k;
}dp[MAXN][3];
int fap[MAXN];
int ans=0;

void dfs(int x,int fa){
	for(int i=0;i<vec[x].size();i++){
		if(vec[x][i]!=fa) dfs(vec[x][i],x);
	}
	for(int i=0;i<vec[x].size();i++){
		int son=vec[x][i];
		if(son==fa) continue;
		if(dp[x][0].k<dp[son][0].k+len[son]) dp[x][0].k=dp[son][0].k+len[son],dp[x][0].Son=son;
	}
	for(int i=0;i<vec[x].size();i++){
		int son=vec[x][i];
		if(son==fa||son==dp[x][0].Son) continue;
		if(dp[x][1].k<dp[son][0].k+len[son]) dp[x][1].k=dp[son][0].k+len[son];
	}
}
void dfs2(int x,int fa){
	if(fa!=-1){
	    dp[x][2].k=dp[fap[x]][2].k+len[x];
		if(dp[fap[x]][0].Son!=x) dp[x][2].k=max(dp[x][2].k,dp[fap[x]][0].k+len[x]);
		else dp[x][2].k=max(dp[x][2].k,dp[fap[x]][1].k+len[x]);
	}
	for(int i=0;i<vec[x].size();i++){
		if(vec[x][i]!=fa) dfs2(vec[x][i],x);
	}
	return ;
}

int main(){
	while(scanf("%d",&N)!=EOF){
		for(int i=1;i<=N;i++){
			dp[i][0].k=dp[i][1].k=dp[i][2].k=0;
			dp[i][0].Son=dp[i][1].Son=0;
			vec[i].clear(); fap[i]=0;
			len[i]=0;
		}
		for(int i=2;i<=N;i++){
			int fat=read(),ltg=read();
			vec[fat].push_back(i);
			len[i]=ltg; fap[i]=fat;
		}
		dfs(1,-1);dfs2(1,-1);
		for(int i=1;i<=N;i++) cout<<max(dp[i][0].k,dp[i][2].k)<<endl;
	}
}

【樹形dp】Computer