紫书动规 P282的问题 hdu2196 树形dp

题目链接:

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

题意:

题解:

http://blog.csdn.net/shuangde800/article/details/9732825
f[i][0],表示顶点为i的子树的，距顶点i的最长距离
f[i][1],表示Tree(i的父节点)-Tree(i)的最长距离+i跟i的父节点距离

要求所有的f[i][0]很简单，只要先做一次dfs求每个结点到叶子结点的最长距离即可。
然后要求f[i][1], 可以从父节点递推到子节点，

代码:

#include <bits/stdc++.h>using namespace std;typedef long long ll;#define MS(a) memset(a,0,sizeof(a))#define MP make_pair#define PB push_backconst int INF = 0x3f3f3f3f;const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}//////////////////////////////////////////////////////////////////////////const int maxn = 1e5+10;struct node{    int v,w;};vector<node> g[maxn];int vis[maxn],f[maxn][2];int dfs1(int u){    vis[u] = 1;    f[u][0] = 0;    for(int i=0; i<(int)g[u].size(); i++){        int v = g[u][i].v, w = g[u][i].w;        if(vis[v]) continue;        f[u][0] = max(f[u][0],dfs1(v)+w);    }    return f[u][0];}void dfs2(int u){    vis[u] = 1;    int max1=0,max2=0,v1,v2;    for(int i=0; i<(int)g[u].size(); i++){        int v = g[u][i].v, w = g[u][i].w;        if(vis[v]) continue; // 不能返回到父节点        int tmp = f[v][0]+w;        if(tmp > max1){            max2 = max1; v2 = v1;            max1 = tmp; v1 = v;        }else if(tmp > max2){            max2 = tmp; v2 = v;        }    }    if(u != 1){        int tmp = f[u][1];        int v = -1;        if(tmp > max1){            max2 = max1; v2 = v1;            max1 = tmp; v1 = v;        }else if(tmp > max2){            max2 = tmp; v2 = v;        }    }    for(int i=0; i<(int)g[u].size(); i++){        int v = g[u][i].v, w = g[u][i].w;        if(vis[v]) continue;        if(v == v1)            f[v][1] = max2 + w;        else            f[v][1] = max1 + w;        // cout << v << " " << f[v][1] << endl;        dfs2(v);    }}int main(){    int n;    while(scanf("%d",&n) == 1){        for(int i=0; i<=n; i++) g[i].clear();        for(int i=2; i<=n; i++){            int v,w; cin >> v >> w;            g[i].push_back(node{v,w});            g[v].push_back(node{i,w});        }        MS(f);        MS(vis);        dfs1(1);        MS(vis);        dfs2(1);        for(int i=1; i<=n; i++)            cout << max(f[i][0],f[i][1]) << endl;    }    return 0;}