HDU-2196 Computer 树形DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2196

　　题意：给一颗树求距离每个节点的最远距离。。

　　每个节点维护两个信息就可以了，f[u][0]和f[u][1]，分别表示子树的最大深度和次大深度，然后一边DFS合并子树就行了。。

//STATUS:C++_AC_15MS_1016KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=10010;
const int INF=0x3f3f3f3f;
const LL MOD=1000000007,STA=8000010;
const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e50;
const int dx[8]={-1,-1,0,1,1,1,0,-1};
const int dy[8]={0,1,1,1,0,-1,-1,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

struct Edge{
    int u,v,w;
}e[N<<1];
int first[N],next[N<<1];
LL f[N][2],ans[N];
int n,mt;

void adde(int a,int b,int c)
{
    e[mt].u=a,e[mt].v=b,e[mt].w=c;
    next[mt]=first[a];first[a]=mt++;
    e[mt].u=b,e[mt].v=a,e[mt].w=c;
    next[mt]=first[b];first[b]=mt++;
}

void dfs1(int u,int fa)
{
    int i,v;
    LL t;
    f[u][0]=f[u][1]=0;
    for(i=first[u];i!=-1;i=next[i]){
        if((v=e[i].v)==fa)continue;
        dfs1(v,u);
        t=f[v][0]+(LL)e[i].w;
        if(t>f[u][0])f[u][1]=f[u][0],f[u][0]=t;
        else if(t>f[u][1])f[u][1]=t;
    }
}

void dfs2(int u,int fa,LL max)
{
    int i,v;
    ans[u]=Max(f[u][0],max);
    for(i=first[u];i!=-1;i=next[i]){
        if((v=e[i].v)==fa)continue;
        if(f[v][0]+e[i].w==f[u][0])
            dfs2(v,u,Max(max,f[u][1])+(LL)e[i].w);
        else dfs2(v,u,Max(max,f[u][0])+(LL)e[i].w);
    }
}

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,a,b,ca=1;
    while(~scanf("%d",&n))
    {
        mem(first,-1);mt=0;
        for(i=2;i<=n;i++){
            scanf("%d%d",&a,&b);
            adde(i,a,b);
        }

        dfs1(1,0);
        dfs2(1,0,0);

        for(i=1;i<=n;i++)
            printf("%I64d
",ans[i]);
    }
    return 0;
}