【树的直径】 POJ 1985 Cow Marathon

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给出一棵树，和边的权值

求权值最长的一条直径

两次bfs求

第一次以任意点开始 BFS求出第一个端点

第二次以第一次得到的端点 BFS求出第二个端点

#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|1typedef pair<int, int> PI;typedef pair<int, PI> PP;#ifdef _WIN32#define LLD "%I64d"#else#define LLD "%lld"#endifconst int MAXN = 200100;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));}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,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;    while(scanf("%d%d",&n,&m)!=EOF)    {        init();        for(int i=0; i<m; i++)        {            scanf("%d%d%d %*c",&a,&b,&c);            addedge(a,b,c);        }        int u=bfs(1);        int v=bfs(u);        int len=d[v];        cout<<len<<endl;    }    return  0;}

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