4
2 1 3 3
1 2 3
1 3 4
2 4 6

 

1
9

 

考慮不換根怎麽做， 那麽每條邊經過的次數就是它的子樹大小。

如果換根， 那麽考慮root的兒子x， 它的答案中只有root到x的邊的貢獻發生變化了。

發生的變化是減去root到x的貢獻， 加上x到root的貢獻。

於是求出root為1的答案， 然後再換根， 這樣統計答案是O(1)的， 總共是O(N)的。

註意更新的順序一定要按照dfs順序， 還要開Long long.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#define int long long
using namespace std;
inline int read() {
    int res=0;char c=getchar();bool f=0;
    while(!isdigit(c)) {if(c==‘-‘)f=1;c=getchar();}
    while(isdigit(c))res=(res<<3)+(res<<1)+(c^48),c=getchar();
    return f?-res:res;
}

#define reg register
#define N 700005

int n;
struct edge {
    int nxt, to, val;
}ed[N*2];
int head[N], cnt;
int ine[N];

inline void add(int x, int y, int z)
{
    ed[++cnt] = (edge){head[x], y, z};
    head[x] = cnt;
}

int a[N], siz[N], fat[N];
int ans, root = 1, res;
int f[N];

void dfs(int x, int fa)
{
    siz[x] = 1;
    for (reg int i = head[x] ; i ; i = ed[i].nxt)
    {
        int to = ed[i].to;
        if (to == fa) continue;
        fat[to] = x;
        ine[to] = i;
        dfs(to, x);
        siz[x] += siz[to];
    }
}

int efs(int x, int fa, int in)
{
    int sum = 0;
    for (reg int i = head[x] ; i ; i = ed[i].nxt)
    {
        int to = ed[i].to;
        if (to == fa) continue;
        sum += efs(to, x, i);
    }
    sum += siz[x] * (ed[in].val - a[fa]);
    return sum;
}

void ffs(int x, int fa, int in)
{
    if (x != 1) f[x] = f[fa] - (siz[x] * (ed[in].val - a[fa])) + (n - siz[x]) * (ed[in].val - a[x]);
    for (reg int i = head[x] ; i ; i = ed[i].nxt)
    {
        int to = ed[i].to;
        if (to == fa) continue;
        ffs(to, x, i);
    }
}

signed main()
{
    freopen("yggdrasil.in", "r", stdin);
    freopen("yggdrasil.out", "w", stdout);
    n = read();
    for (reg int i = 1 ; i <= n ; i ++) a[i] = read();
    for (reg int i = 1 ; i < n ; i ++)
    {
        int x = read(), y = read(), z = read();
        add(x, y, z);add(y, x, z);
    }
    dfs(1, 0);
    res = efs(1, 0, 0);
    ans = res;
    f[1] = ans;
    ffs(1, 0, 0);
    for (reg int i = 2 ; i <= n ; i ++) 
        if (f[i] < ans) {ans = f[i], root = i;}
    printf("%lld\n%lld\n", root, ans);
    return 0;
} 

[JZOJ5776]【NOIP2008模擬】小x遊世界樹