HDU 2196 Computer 經典樹形DP

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一開始看錯題了，後來發現原來是在一顆帶權的樹上面求出距離每一個點的最長距離，做兩次dfs就好，具體的看注釋？

#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求解從子節點過來的最大值和次大值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) {    //如果當前節點是葉子節點，最大值必定是從父親節點更新過來的    if(ch[now] == -1) return;    //否則往下更新    for(int i = ch[now];~i;i = nxt[i]) {        if(mfrom[now] == v[i]) {            //如果當前兒子這邊有最大深度            f[v[i]] = max(f[now],sdep[now]) + w[i];        }        else f[v[i]] = max(f[now],mdep[now]) + w[i];        dfs1(v[i]);    }    //最大距離來自兒子    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;}