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。。
  1 //STATUS:C++_AC_15MS_1016KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 #pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef __int64 LL;
 34 typedef unsigned __int64 ULL;
 35 //const
 36 const int N=10010;
 37 const int INF=0x3f3f3f3f;
 38 const LL MOD=1000000007,STA=8000010;
 39 const LL LNF=1LL<<55;
 40 const double EPS=1e-9;
 41 const double OO=1e50;
 42 const int dx[8]={-1,-1,0,1,1,1,0,-1};
 43 const int dy[8]={0,1,1,1,0,-1,-1,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 struct Edge{
 59     int u,v,w;
 60 }e[N<<1];
 61 int first[N],next[N<<1];
 62 LL f[N][2],ans[N];
 63 int n,mt;
 64 
 65 void adde(int a,int b,int c)
 66 {
 67     e[mt].u=a,e[mt].v=b,e[mt].w=c;
 68     next[mt]=first[a];first[a]=mt++;
 69     e[mt].u=b,e[mt].v=a,e[mt].w=c;
 70     next[mt]=first[b];first[b]=mt++;
 71 }
 72 
 73 void dfs1(int u,int fa)
 74 {
 75     int i,v;
 76     LL t;
 77     f[u][0]=f[u][1]=0;
 78     for(i=first[u];i!=-1;i=next[i]){
 79         if((v=e[i].v)==fa)continue;
 80         dfs1(v,u);
 81         t=f[v][0]+(LL)e[i].w;
 82         if(t>f[u][0])f[u][1]=f[u][0],f[u][0]=t;
 83         else if(t>f[u][1])f[u][1]=t;
 84     }
 85 }
 86 
 87 void dfs2(int u,int fa,LL max)
 88 {
 89     int i,v;
 90     ans[u]=Max(f[u][0],max);
 91     for(i=first[u];i!=-1;i=next[i]){
 92         if((v=e[i].v)==fa)continue;
 93         if(f[v][0]+e[i].w==f[u][0])
 94             dfs2(v,u,Max(max,f[u][1])+(LL)e[i].w);
 95         else dfs2(v,u,Max(max,f[u][0])+(LL)e[i].w);
 96     }
 97 }
 98 
 99 int main(){
100  //   freopen("in.txt","r",stdin);
101     int i,j,a,b,ca=1;
102     while(~scanf("%d",&n))
103     {
104         mem(first,-1);mt=0;
105         for(i=2;i<=n;i++){
106             scanf("%d%d",&a,&b);
107             adde(i,a,b);
108         }
109 
110         dfs1(1,0);
111         dfs2(1,0,0);
112 
113         for(i=1;i<=n;i++)
114             printf("%I64d
",ans[i]);
115     }
116     return 0;
117 }