HDU 2196 Computer (树形DP)

题意：给定一棵树，然后让你找出每个结点离所有结点的最远距离。

析：也就说我们要知道离每个结点的最远距离，对于每个结点，我们知道离它最远的，要么是从父结点过来，要么是从子树中得到，dp[i][0] 表示从 i 子树中得到的

最远距离，dp[i][1] 表示 i 从子树得到的次远距离，dp[i][2] 表示从 父结点得到的最大距离，我们要搜索两次，第一次就是搜索子树，很容易得到最远距离和次远距离，

第二次就是要计算从父结点得到的距离，这个要用到这个次远距离，因为我们不知道次远距离和从父结点来的哪个更大。最后再比较dp[i][0] 和 dp[i][2] 就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
    int to, val, next;
};
Edge edge[maxn<<1];
int head[maxn], dp[maxn][3];
int cnt;

void add(int u, int v, int val){
    edge[cnt].to = v;
    edge[cnt].val = val;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}

void dfs(int u, int fa){
    int mmax = 0, lmax = 0;
    for(int i = head[u]; ~i; i = edge[i].next){
        int v = edge[i].to;
        int val = edge[i].val;
        if(v == fa)  continue;
        dfs(v, u);
        int tmp = dp[v][0] + val;
        if(lmax <= tmp){
            mmax = lmax;
            lmax = tmp;
        }
        else if(mmax < tmp)  mmax = tmp;
    }
    dp[u][0] = lmax;
    dp[u][1] = mmax;
}

 void dfs1(int u, int fa){
    for(int i = head[u]; ~i; i = edge[i].next){
        int v = edge[i].to;
        int val = edge[i].val;
        if(v == fa)  continue;
        dp[v][2] = Max(dp[u][2], dp[v][0]+val == dp[u][0] ? dp[u][1] : dp[u][0]) + val;
        dfs1(v, u);
    }
 }

int main(){
    while(scanf("%d", &n) == 1){
        int u, c;
        memset(head, -1, sizeof head);
        cnt = 0;
        for(int i = 2; i <= n; ++i){
            scanf("%d %d", &u, &c);
            add(i, u, c);
            add(u, i, c);
        }
        memset(dp, 0, sizeof dp);
        dfs(1, -1);
        dfs1(1, -1);
        for(int i = 1; i <= n; ++i)
            printf("%d\n", Max(dp[i][0], dp[i][2]));
    }
    return 0;
}