历届试题 大臣的旅费

树的直径，利用树形dp

待补dfs*2的方法

#include<bits/stdc++.h>
using namespace std;
#define fore(i,a,b) for(int i=a;i<=b;i++)
#define forb(i,a,b) for(int i=a;i>=b;i--)
#define fir first
#define sec second
#define ABS(a) ((a)>0?(a):-(a))
#define PI acos(-1,0)
typedef long long ll;
typedef long double ld;
const int MAXN=100005;
int ver[MAXN*10],Next[MAXN],edge[MAXN],head[MAXN];
bool vis[MAXN];
int tot;
int ans;
int d[MAXN],f[MAXN];

void add(int x,int y,int z){
    ver[++tot]=y,edge[tot]=z;
    Next[tot]=head[x],head[x]=tot;
    ver[++tot] = x, edge[tot] = z;
    Next[tot] = head[y], head[y] = tot;
}

void dp(int x){
    vis[x]=true;
    for(int i=head[x];i;i=Next[i]){
        int y=ver[i];
        if(vis[y]) continue;
        dp(y);
        ans=max(ans,d[x]+d[y]+edge[i]);
        d[x] = max(d[x], d[y] + edge[i]);
    }
}

int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        add(x,y,z);
    }
    dp(1);
    cout<<(ans*10+(ans+1)*ans/2);
    return 0;
}

标签：head,试题,int,tot,edge,MAXN,旅费,历届,define
来源： https://www.cnblogs.com/rign/p/10472660.html