bzoj1509

树的直径

我先开始以为是个图，想想并不知道什么求图的直径的方法，结果是棵树

那么直觉告诉我们是在直径上面，实际上就是直径+min(i->u,i->v)，扫一遍就行了

#include<bits/stdc++.h>
using namespace std;
const int N = 200010;
namespace IO 
{
    const int Maxlen = N;
    char buf[Maxlen], *C = buf;
    int Len;
    inline void read_in()
    {
        Len = fread(C, 1, Maxlen, stdin);
        buf[Len] = '\0';
    }
    inline void fread(int &x) 
    {
        x = 0;
        int f = 1;
        while (*C < '0' || '9' < *C) { if(*C == '-') f = -1; ++C; }
        while ('0' <= *C && *C <= '9') x = (x << 1) + (x << 3) + *C - '0', ++C;
        x *= f;
    }
    inline void read(int &x)
    {
        x = 0;
        int f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = (x << 1) + (x << 3) + c - '0'; c = getchar(); }
        x *= f;
    }
    inline void read(long long &x)
    {
        x = 0;
        long long f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = (x << 1ll) + (x << 3ll) + c - '0'; c = getchar(); }
        x *= f;
    } 
} using namespace IO;
struct edge {
    int to;
    long long w;
    edge(int to = 0, long long w = 0) : to(to), w(w) {}
};
int n, m, root;
long long ans;
vector<edge> G[N];
long long f[N], g[N], d[N];
void dfs(int u, int last, long long d[])
{
    if(d[u] > d[root]) root = u;
    for(int i = 0; i < G[u].size(); ++i)
    {
        edge e = G[u][i];
        if(e.to == last) continue;
        d[e.to] = d[u] + e.w;
        dfs(e.to, u, d);
    }    
}
int main()
{
    read(n);
    read(m);
    for(int i = 1; i <= m; ++i)
    {
        int u, v; 
        long long w;
        read(u);
        read(v);
        read(w);
        G[u].push_back(edge(v, w));
        G[v].push_back(edge(u, w));
    }
    dfs(1, 0, f);
    int u = root;
    root = 0;
    dfs(u, 0, g);
    long long dis = g[root];
    u = root;
    root = 0;
    dfs(u, 0, d);
    for(int i = 1; i <= n; ++i) ans = max(ans, dis + min(d[i], g[i]));
    printf("%lld\n", ans);
    return 0;
}