【树的直径】 POJ 2631 Roads in the North

转自 DISCUSS的证明

设最长链是MN->已知[1]

设由B开始DFS得到最长路为BC->已知[2]

结论[1] MN与AB有公共点.否则MN<AM+AN<=AM+AB=BM 与已知[1]矛盾

结论[2] B是最长链的一个端点.否则由结论[1] 设K是AB上距B最近且在MN上的点 则MN=MK+KN=MK+AN-AK<=MK+AB-AK=MK+BK=BM 当取等号时MB与MN等长 符合结论[2] 否则与已知[1]矛盾 [这里假定了A不在NK上.若A在NK上 只须将上面式子中MN交换位置即可 不影响结论]

结论[3] BC是一条最长链.否则由结论[2] 设经过B的最长链为BD 则BD>BC 与已知[2]矛盾

至此证毕

（数据居然可能为空。。。）

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <cctype>
#include <cmath>
#include <string>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
typedef long long LL;
typedef long double LD;
#define pi acos(-1.0)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
typedef pair<int, int> PI;
typedef pair<int, PI> PP;
#ifdef _WIN32
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif
const int MAXN = 20100;
const int INF = 999999;
//LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
//inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}
//inline void print(LL x){printf(LLD, x);puts("");}
//inline void read(int &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}
struct Edge
{
    int to,next,val;
} edge[MAXN*2];
int head[MAXN],d[MAXN],tol;
bool vis[MAXN];
void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
    memset(d,0,sizeof(d));
}
void addedge(int u,int v,int w)
{
    edge[tol].to=v,edge[tol].val=w,edge[tol].next=head[u];
    head[u]=tol++;
    edge[tol].to=u,edge[tol].val=w,edge[tol].next=head[v];
    head[v]=tol++;
}
int bfs(int u)
{
    int point=0,big=0;
    memset(vis,false,sizeof(vis));
    queue<int>q;
    q.push(u);
    vis[u]=true;
    d[u]=0;
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        for(int i=head[u]; ~i; i=edge[i].next)
        {
            int v=edge[i].to;
            if(!vis[v])
            {
                d[v]=d[u]+edge[i].val;
                if(d[v]>big)
                {
                    big=d[v];
                    point=v;
                }
                vis[v]=true;
                q.push(v);
            }
        }
    }
    return point;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //  freopen("out.txt", "w", stdout);
#endif
    int t,a,b,c,n,m,len=0;
    init();
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        addedge(a,b,c);
    }
    int u=bfs(1);
    int v=bfs(u);
    len=d[v];
    cout<<len<<endl;
    return 0;
}