1509: [NOI2003]逃学的小孩

Description

Input

第一行是两个整数N（3 < N < 200000）和M，分别表示居住点总数和街道总数。以下M行，每行给出一条街道的信息。第i+1行包含整数Ui、Vi、Ti（1 < Ui, Vi < N，1 < Ti < 1000000000），表示街道i连接居住点Ui和Vi，并且经过街道i需花费Ti分钟。街道信息不会重复给出。

Output

仅包含整数T，即最坏情况下Chris的父母需要花费T分钟才能找到Chris。

Sample Input

4 3
1 2 1
2 3 1
3 4 1

Sample Output

4

打完代码，感觉身体被掏空，不多说了，直接上图。

大概就是这个感觉，恩恩。
代码如下

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int sz = 2001000;
int head[sz],nxt[sz];
ll dis_s[sz],dis_t[sz];
int tot = 1;
int n,m;
struct gtnd
{
    int t;
    ll d;
}l[sz];
void build_edge(int f,int t,ll d)
{
    l[tot].t = t;
    l[tot].d = d;
    nxt[tot] = head[f];
    head[f] = tot ++;
}
int read_int()
{
    int x = 0 , f = 1;
    char in = getchar();
    while(in < '0' || in > '9')
    {
        if(in == '-')
            f = -1;
        in = getchar();
    }
    while('0' <= in && in <= '9')
    {
        x = (x << 3) + (x << 1) + in - '0';
        in = getchar();
    }
    return x * f;
}
ll read_ll()
{
    ll x = 0 , f = 1;
    char in = getchar();
    while(in < '0' || in > '9')
    {
        if(in == '-')
            f = -1;
        in = getchar();
    }
    while('0' <= in && in <= '9')
    {
        x = x * 10 + in - '0';
        in = getchar();
    }
    return x * f;
}
void start_work()
{
    n = read_int() , m = read_int();
    for(int i = 1 ; i <= m ; i ++)
    {
        int f = read_int() , t = read_int();
        ll d = read_ll();
        build_edge(f,t,d);
        build_edge(t,f,d);
    }
}
int pos;
ll dist;
void dfs_first(int u,int fa,ll d)
{
    if(d > dist)
        dist = d , pos = u;
    for(int i = head[u] ; i ; i = nxt[i])
        if(l[i].t != fa)
            dfs_first(l[i].t,u,d+l[i].d);
}
void dfs_s(int u,int fa,ll d)
{
    dis_s[u] = d;
    for(int i = head[u] ; i ; i = nxt[i])
        if(l[i].t != fa)
            dfs_s(l[i].t,u,d+l[i].d);
}
void dfs_t(int u,int fa,ll d)
{
    dis_t[u] = d;
    for(int i = head[u] ; i ; i = nxt[i])
        if(l[i].t != fa)
            dfs_t(l[i].t,u,d+l[i].d);
}
int main()
{
    start_work();
    dfs_first(1,0,0);
    int s = pos;
    dist = 0;
    dfs_first(pos,0,0);
    int t = pos;
    dfs_s(s,0,0);
    dfs_t(t,0,0);
    ll ans = 0;
    for(int i = 1 ; i <= n ; i ++)
        if(i != s && i != t)
            ans = max(ans,min(dis_s[i],dis_t[i]));
    printf("%lld\n",ans+dist);
    return 0;
}