Graph Theory for sdut3045-(Multi-tree longest chain) and sdut3045 Graph Theory

Graph Theory for sdut3045-(Multi-tree longest chain) and sdut3045 Graph Theory

Fan Zhi Graph Theory Time Limit: 1000 MS Memory limit: 65536 K Description

FF is a master of graph theory, so I want to figure out a graph without any flow problems.

Returns the length of the longest chain of a tree.

Multiple Input groups. The first Act n (1 <= n <= 100000) entered in each group represents the number of nodes. The node number ranges from 1 to n. The next n-1 rows have two positive integers u, v, represents an edge connecting u and v. Make sure that each group of data is a tree.

For each group of data output, a positive integer represents the answer.

Sample Input

121 2

Sample output

12

Prompt source zmxdfs search

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;struct node{    int v ;    int next ;}p[1000000];int head[110000] , cnt , vis[110000] ;int ans , max1[110000] ;void add(int u,int v){    p[cnt].v = v ;    p[cnt].next = head[u] ; head[u] = cnt++ ;    p[cnt].v = u ;    p[cnt].next = head[v] ; head[v] = cnt++ ;}void dfs(int u,int num){    int i , j , v , flag = 0 , k1 = -1 , k2 = -1 ;    for(i = head[u] ; i != -1 ; i = p[i].next)    {        v = p[i].v ;        if( !vis[v] )        {            flag = 1 ;            vis[v] = num+1 ;            dfs(v,num+1) ;        }    }    if( flag == 0 )    {        max1[u] = 1 ;        return ;    }    for(i = head[u] ; i != -1 ; i = p[i].next)    {        v = p[i].v ;        if( vis[u]+1 != vis[v] )            continue ;        if( k1 == -1 )            k1 = max1[v] ;        else        {            if( max1[v] > k1 )            {                k2 = k1 ;                k1 = max1[v] ;            }            else if(  max1[v] > k2 )                k2 = max1[v] ;        }    }    max1[u] = k1 + 1 ;    ans = max( ans,max1[u] ) ;    if( k2 == -1 )        return ;    else    {        ans = max( ans,k1+k2+1 ) ;    }    return ;}int main(){    int n , u , v , i ;    while( scanf("%d", &n) != EOF )    {        cnt = 0 ;        ans = 2 ;        memset(head,-1,sizeof(head)) ;        memset(vis,0,sizeof(vis)) ;        if(n == 1)        {            printf("1\n") ;            continue ;        }        for(i = 1 ; i < n ; i++)        {            scanf("%d %d", &u, &v) ;            add(u,v) ;        }        vis[1] = 1 ;        dfs(1,1);        printf("%d\n", ans) ;    }    return 0;}