BZOJ 1509 逃学的小孩(树的直径)

题意:从树上任找三点u，v，w。使得dis(u,v)+min(dis(u,w),dis(v,w))最大。

 

有一个结论u，v必是树上直径的两端点。 剩下的枚举w就行了。

具体不会证。。。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=200005;
//Code begin...

struct Edge{int p, next, w;}edge[N<<1];
int head[N], cnt=1;
LL dis[N], dist[N];

void add_edge(int u, int v, int w){
    edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++;
}
void dfs(int x, int fa){
    for (int i=head[x]; i; i=edge[i].next) {
        int v=edge[i].p;
        if (v==fa) continue;
        dis[v]=dis[x]+edge[i].w;
        dfs(v,x);
    }
}
int main ()
{
    int n, m, u, v, w;
    LL ans=0;
    scanf("%d%d",&n,&m);
    while (m--) scanf("%d%d%d",&u,&v,&w), add_edge(u,v,w), add_edge(v,u,w);
    dfs(1,0);
    int ma=0;
    FOR(i,1,n) if(dis[ma]<dis[i]) ma=i;
    mem(dis,0); dfs(ma,0);
    ma=0;
    FOR(i,1,n) {
        dist[i]=dis[i];
        if (dis[ma]<dis[i]) ma=i;
    }
    mem(dis,0); dfs(ma,0);
    FOR(i,1,n) ans=max(ans,dist[ma]+min(dis[i],dist[i]));
    printf("%lld\n",ans);
    return 0;
}