洛谷_P13099 [FJCPC 2025] VERTeX

题目链接：P13099 [FJCPC 2025] VERTeX - 洛谷

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题目大意：

给定一棵树，边权等于两端点权之和，已知边权

判断是否存在正整数点权并构造。

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思路：

首先，如果某个节点的权值确定，那么所有节点的权值也都确定

我们从某个节点入手，假设该节点的权值为1，

得出所有节点权值，若该方案合法，直接输出即可

若不合法，说明必定存在某个节点的权值是非正数，

我们调整该节点的权值，让所有节点权值变为正数

再次判断即可

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代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<bitset>
#include<tuple>
#include<array>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define mst(a,x) memset(a,x,sizeof (a))
using namespace std;
typedef pair<int, int> pii;

const int N = 400086, mod = 998244353;

int n, m;
int h[N], ne[N], e[N], w[N], idx;
int res[N];

void add(int a, int b, int c) {
    w[idx] = c, e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void dfs(int u, int fa) {
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j == fa) continue;
        res[j] = w[i] - res[u];
        dfs(j, u);
    }
}

void solve() {

    mst(h, -1);
    cin >> n;
    for (int i = 1; i < n; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c), add(b, a, c);
    }
    
    res[1] = 1;
    dfs(1, 0);
    int mn = *min_element(res + 1, res + n + 1);
    
    if (mn <= 0) {
        fill(res, res + n + 1, 0), res[1] = 2 - mn;
        dfs(1, 0);
        mn = *min_element(res + 1, res + n + 1);
        cout << (mn <= 0 ? "NO" : "YES") << endl;
    } else cout << "YES" << endl;
    
    if (mn > 0) for (int i = 1; i <= n; i++) cout << res[i] << ' ';
    
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
// cin >> T;
    while (T--) solve();
    
    return 0;
}