HDU 2196 Computer 经典树形DP

一开始看错题了，后来发现原来是在一颗带权的树上面求出距离每一个点的最长距离，做两次dfs就好，具体的看注释？

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#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack>   using namespace std;   typedef long long LL; const int maxn = 10000 + 5; int ch[maxn],nxt[maxn],v[maxn],w[maxn]; int mdep[maxn],sdep[maxn],f[maxn],mfrom[maxn]; int ecnt,n;   inline void addedge(int uu,int vv,int ww) {     v[ecnt] = vv; w[ecnt] = ww; nxt[ecnt] = ch[uu]; ch[uu] = ecnt++; }   //dfs求解从子节点过来的最大值和次大值 int dfs(int now) {     mdep[now] = sdep[now] = 0;     if(ch[now] == -1) return 0;     for(int i = ch[now];~i;i = nxt[i]) {         int ret = dfs(v[i]) + w[i];         if(ret > mdep[now]) {             sdep[now] = mdep[now]; mdep[now] = ret;             mfrom[now] = v[i];         }         else if(ret > sdep[now]) sdep[now] = ret;     }     return mdep[now]; }   //dfs1求出答案 void dfs1(int now) {     //如果当前节点是叶子节点，最大值必定是从父亲节点更新过来的     if(ch[now] == -1) return;     //否则往下更新     for(int i = ch[now];~i;i = nxt[i]) {         if(mfrom[now] == v[i]) {             //如果当前儿子这边有最大深度             f[v[i]] = max(f[now],sdep[now]) + w[i];         }         else f[v[i]] = max(f[now],mdep[now]) + w[i];         dfs1(v[i]);     }     //最大距离来自儿子     f[now] = max(f[now],mdep[now]); }   int main() {     while(scanf("%d",&n) != EOF) {         memset(ch,-1,sizeof(ch));         memset(f,0,sizeof(f));         memset(mfrom,-1,sizeof(mfrom));         ecnt = 0;         for(int i = 2;i <= n;i++) {             int a,b; scanf("%d%d",&a,&b);             addedge(a,i,b);         }         dfs(1);         dfs1(1);         for(int i = 1;i <= n;i++) printf("%d\n",f[i]);     }     return 0; }