AC绿题整理

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抽空a寄到绿题(bushi)

P1131ZJOI2007

首先可以感性证明这是一颗树, 发现可以贪心地做, 记录一个max, 当子树中的每一个叶子节点都需要补全的时候最优在子树的根与他父亲的边增加长度, 附上代码.

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#include <bits/stdc++.h>
using namespace std;

#define int long long
#define maxn 500001
#define rep(i, a, b) for(int i = (a); i <= (b); i ++)
#define inf 0x3f3f3f3f

int n, s;

int leaf[maxn], tim;

int head[maxn], nxt[maxn<<1], dat[maxn<<1], cnt, val[maxn<<1];
void add(int u, int v, int w)
{
    dat[++ cnt] = v; nxt[cnt] = head[u]; head[u] = cnt; val[cnt] = w;
}

int lg[maxn];

long long ans = 0;

int dfs(int x, int fa)
{
    int cnt = 0;
    int mx = -inf;
    int sum = 0; 
    for(int i = head[x]; i; i = nxt[i])
    { 
        int da = dat[i];
        if(da == fa) continue; 
        lg[da] = lg[x] + val[i];
        int t = dfs(da, x);
        // if(x == 4) cout << t << endl;
        sum += t;
        mx = max(mx, t);
        
        cnt ++;
    }
    if(mx == -inf) mx = lg[x];
    ans += mx*cnt - sum;
    // cout << x << " " << mx << " " << cnt << " " << sum << " " << mx*cnt - sum << endl;
    return mx;
}

signed main()
{
    freopen("a.in", "r", stdin);
    freopen("a.out", "w", stdout);
    cin >> n >> s;
    rep(i, 1, n - 1)
    {
        int u, v, w; cin >> u >> v >> w;
        add(u, v, w);
        add(v, u, w);
    }

    dfs(s, 0); 
    cout << ans << endl;
}