[hihocoder#1050 : 树中的最长路] 两种树形DP方法求树的最长路
方法一:以树上任意一个节点为根节点出发求出他子树的最长距离,那么距离最大的那个节点就必然是最长路的一个端点。然后以这个端点为根节点再DFS一次就可以求出答案了。需要DFS两次。
方法二:这个思路就是按照hihocoder 的提示思路一样,取转折点,从任意一个节点为根节点出发,遍历整个树,求出每个节点所在子树到其本身的最长路和次长路。然后,答案就是最长路+次长路了。
/**
* 方法一:两次DFS求端点
*/
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define fst first
#define snd second
//typedef __int64 LL;
typedef long long LL;
typedef pair<int, int> PII;
const double eps = 1e-6;
const int MAXN = 1e5 + 5;
struct Edge {
int v, next;
Edge() {}
Edge(int v, int next): v(v), next(next) {}
} edges[MAXN << 1];
int head[MAXN], tot, N;
PII rec;
void init() {
tot = 0;
rec = PII(-1, -1);
memset(head, -1, sizeof(head));
}
void add_edge(int u, int v) {
edges[tot] = Edge(v, head[u]);
head[u] = tot ++;
}
void dfs(int u, int pre, int d) {
int v;
if(rec.fst < d) {
rec.fst = d;
rec.snd = u;
}
for (int i = head[u]; ~i; i = edges[i].next) {
v = edges[i].v;
if (v == pre) continue;
dfs(v, u, d + 1);
}
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
int u, v;
while (~scanf("%d", &N)) {
init();
for (int i = 1; i < N; i++) {
scanf("%d %d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
dfs(1, -1, 0);
dfs(rec.snd, -1, 0);
printf("%d\n", rec.fst);
}
return 0;
}
/**
* 方法二:一次DFS求转折点
*/
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define fst first
#define snd second
//typedef __int64 LL;
typedef long long LL;
typedef pair<int, int> PII;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 5;
struct Edge {
int v, next;
Edge() {}
Edge(int v, int next): v(v), next(next) {}
} edges[MAXN << 1];
int head[MAXN], tot, N, res;
int dp[MAXN][2];
void init() {
tot = 0;
res = -INF;
memset(head, -1, sizeof(head));
}
void add_edge(int u, int v) {
edges[tot] = Edge(v, head[u]);
head[u] = tot ++;
}
void dfs(int u, int pre) {
int v;
dp[u][0] = dp[u][1] = 0;
for (int i = head[u]; ~i; i = edges[i].next) {
v = edges[i].v;
if (v == pre) continue;
dfs(v, u);
int temp = dp[v][1] + 1;
if (temp > dp[u][1]) swap(temp, dp[u][1]);
if (temp > dp[u][0]) swap(temp, dp[u][0]);
}
res = max(res, dp[u][1] + dp[u][0]);
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
int u, v;
while (~scanf("%d", &N)) {
init();
for (int i = 1; i < N; i++) {
scanf("%d %d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
dfs(1, -1);
printf("%d\n", res);
}
return 0;
}