hiho 1050 树中的最长路 (树的直径)

  最近在复习比较简单的知识，顺便当整理代码吧。

  树的直径是一个经典问题，即求树上最远两点的距离。

思路一：

  任取一个点，求这个点的最远点的最远点，两遍bfs即可。

 代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <functional>
#include <cctype>
#include <time.h>

using namespace std;

const int INF = 1<<30;

struct Graph {
    static const int MAXN = (int) 1e5+55;
    static const int MAXM = MAXN*2;

    int head[MAXN];
    int next[MAXM], to[MAXM], use;

    inline void init() {
        memset(head, -1, sizeof(head));
        use = 0;
    }

    inline void addEdge(int u, int v) {
        next[use] = head[u];
        to[use] = v;
        head[u] = use++;
    }
};

const int MAXN = (int) 1e5+55;

Graph G;
int Que[MAXN];
int d[MAXN];
int n;

void bfs(int st) {
    memset(d, -1, sizeof(d));
    int l = 0, r = 0;
    d[st] = 0;
    Que[r++] = st;
    while (l<r) {
        int u = Que[l++];
        for (int p = G.head[u]; p != -1; p = G.next[p]) {
            int v = G.to[p];
            if (d[v]!=-1) continue;
            d[v] = d[u] + 1;
            Que[r++] = v;
        }
    }
}

void solve() {
    bfs(1);
    bfs(Que[n-1]);
    printf("%d
", d[Que[n-1]]);
}

int main() {
    #ifdef Phantom01
        freopen("1050.txt", "r", stdin);
    #endif //Phantom01

    while (scanf("%d", &n)!=EOF) {
        G.init();
        int u, v;
        for (int i = 1; i < n; i++) {
            scanf("%d%d", &u, &v);
            G.addEdge(u, v);
            G.addEdge(v, u);
        }
        solve();
    }

    return 0;
}

另一种思路是树形dp，一棵树中的最长路要么在子树中，要么经过根。枚举每棵子树的最大深度和次大深度即可。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <functional>
#include <cctype>
#include <time.h>

using namespace std;

const int INF = 1<<30;

struct Graph {
    static const int MAXN = (int) 1e5+55;
    static const int MAXM = MAXN*2;

    int head[MAXN];
    int next[MAXM], to[MAXM], use;

    inline void init() {
        memset(head, -1, sizeof(head));
        use = 0;
    }

    inline void addEdge(int u, int v) {
        next[use] = head[u];
        to[use] = v;
        head[u] = use++;
    }
};

Graph G;

const int MAXN = (int) 1e5+55;

int dp[MAXN];
int ans;

int dfs(int u, int pre, int cnt) {
    int fi = 0, se = 0;
    int t;

    for (int p = G.head[u]; p != -1; p = G.next[p]) {
        int v = G.to[p];
        if (v==pre) continue;
        t = dfs(v, u, cnt+1);
        if (t>fi) { se = fi; fi = t; }
        else if (t>se) se = t;
    }
    ans = max(ans, max(fi+cnt, fi+se));
    return fi+1;
}

int main() {
    #ifdef Phantom01
        freopen("1050.txt", "r", stdin);
    #endif //Phantom01

    int n, u, v;
    while (scanf("%d", &n)!=EOF) {
        G.init();
        for (int i = 1; i < n; i++) {
            scanf("%d%d", &u, &v);
            G.addEdge(u, v);
            G.addEdge(v, u);
        }
        ans = 0;
        dfs(1, 1, 0);
        printf("%d
", ans);
    }

    return 0;
}

顺便说一下，hiho上这个题的数据比较弱，开始我几个错误的代码都A了。