CF1515F Phoenix and Earthquake

https://www.luogu.com.cn/problem/CF1515F

首先必须保证 s u m a [ i ] ≥ ( n − 1 ) ∗ x suma[i]\ge (n-1)*x suma[i]≥(n−1)∗x

可以证明是充要的
考虑随便拉一棵树出来，从叶子结点考虑，如果 a [ u ] ≥ x a[u]\ge x a[u]≥x，那么把 u u u和父亲合并，否则先不管 u u u,把 u u u删了，剩下的又是一个子问题，然后最后再把子问题得到的那个点和 u u u合并

code:

#include<bits/stdc++.h>
#define N 300050
#define ll long long
using namespace std;
struct edge {
    int v, nxt, c;
} e[N << 1];
int p[N], eid;
void init() {
    memset(p, -1, sizeof p);
    eid = 0;
}
void insert(int u, int v, int c) { //printf("%d --> %d  %d\n", u, v, c);
    e[eid].v = v;
    e[eid].nxt = p[u];
    e[eid].c = c;
    p[u] = eid ++;
}
int fa[N];
int get(int x) {
    return fa[x] == x? x : (fa[x] = get(fa[x]));
}
int l, r, n, m, K, ans[N];
ll a[N];
void dfs(int u, int fa, int id) {
    for(int i = p[u]; i + 1; i = e[i].nxt) {
        int v = e[i].v, c = e[i].c;
        if(v == fa) continue;
      //  printf("%d --> %d %d\n", u, v, c);
        dfs(v, u, c);
    }
 //   printf("  %d %d %d %d\n", u, a[u], l, r);
    if(u == 1) return ;
    if(a[u] >= K) a[fa] += a[u] - K, ans[l ++] = id;
    else ans[r --] = id;
}
int main() {
    init();
    scanf("%d%d%d", &n, &m, &K);
    ll s = 0;
    for(int i = 1; i <= n; i ++) fa[i] = i, scanf("%lld", &a[i]), s += a[i];
    if(s < 1ll * (n - 1) * K) {
        printf("NO");
        return 0;
    }
    for(int i = 1; i <= m; i ++) {
        int x, y;
        scanf("%d%d", &x, &y);
        if(get(x) == get(y)) continue;
        insert(x, y, i), insert(y, x, i);
        fa[get(x)] = get(y);
    }
    l = 1, r = n - 1;
    dfs(1, 1, 0);
    //printf("  %d %d\n", l, r);
    printf("YES\n");
    for(int i = 1; i < n; i ++) printf("%d\n", ans[i]);
    return 0;
}

标签：return,Phoenix,fa,CF1515F,get,int,eid,printf,Earthquake
来源： https://blog.csdn.net/qq_38944163/article/details/121327934