Codeforces 1091E

题意：给定n个点的度数，请你添加第n+1个点，输出所有可能的第n+1个点的度数

做法：1. 查看链接知道了下面这个定理

A sequence of non-negative integers $ d_1geqcdotsgeq d_n$ can be represented as the degree sequence of a finite simple graph on (n) vertices if and only if $ d_1+cdots+d_n$ is even and(sum _{i=1}^{k}d_{i}leq k(k-1)+sum _{i=k+1}^{n}min(d_{i},k))
holds for every k in ({displaystyle 1leq kleq n}).

2.排序之后，([low(p), up(p)]) 表示在([p, p+1]) 之间插入新的点度数(d_t)的取值范围。

(1) (d_{p+1} leq d_t leq d_p)

(2) 当 (d_t) 在不等式左侧时

[d_t leq k(k+1)+sum _{i=k+1}^{n}min(d_{i},k+1) - sum _{i=1}^{k}d_{i}, p leq k leq n \ Downarrow \ up(p) = min(up(p + 1) , p(p+1) + sum_{i=p+1}^n min(d_i,p+1) - sum_{i=1}^p d_i) ]

(3) 当 (d_t) 在不等式右侧时

[sum_{i=1}^k d_i - k(k-1) - sum_{i=k+1}^n min(d_i,k) <= d_t , 1 leq k leq p \ Downarrow\ low(p) = max(sum_{i=1}^p d_i - p(p-1) - sum_{i=p+1}^n min(d_i,p), low(p-1)) \ ]

3.猜测答案为连续的一段值，将上一步求出的区间并起来

4.又因为无向简单图，每条边对度数贡献为2，由此可以确定答案的奇偶性

5.确定上下界时，需要维护(sum_{i = k+1}^{n} min(d_i,k)) 和(sum_{i = k}^{n} min(d_i,k)) 我们倒着枚举(k)，用优先队列维护小于等于当前(k)的集合，以及这个集合内所有元素的和，就可以预处理出这个式子，因为k是递减的因此保证了正确性

6.一定要分析清楚再写。。。代码细节很多

#include <bits/stdc++.h>
typedef long long ll;
constexpr int N = 600005;
constexpr ll inf = 1e18;
using namespace std;
int n;
priority_queue<ll> q;
ll f1[N], f2[N], sum[N], low[N], up[N], d[N];
int main() {
    scanf("%d",&n);
    for(int i = 1; i <= n; ++i) scanf("%lld",&d[i]);
    sort(d+1,d+1+n); reverse(d+1,d+1+n);
    for(int i = 1; i <= n; ++i) sum[i] = sum[i-1] + d[i];
    d[0] = n; d[n+1] = 0;
    for(int i = 0; i <= n; ++i) up[i] = inf, low[i] = -inf;
    ll tmp = 0;
    for(int i = n; i >= 0; --i) {
        while(!q.empty() && q.top() > i) tmp -= q.top(), q.pop();
        f2[i] = tmp + 1ll*(n - i - q.size())*i;
        if(d[i] <= i) q.push(d[i]), tmp += d[i];
        f1[i] = tmp + 1ll*(n - i + 1 - q.size())*i;
    }
    for(int i = n; i >= 0; --i) {
        if(i!=n) up[i] = min(up[i+1], 1ll*i*(i+1) + f1[i+1] - sum[i]);
        else up[i] = 1ll*i*(i+1) - sum[i];
    }
    for(int i = 1; i <= n; ++i) {
        low[i] = max(low[i-1], sum[i] - 1ll*i*(i-1) - f2[i]);
    }
    for(int i = 0; i <= n; ++i) low[i] = max(low[i], d[i+1]), up[i] = min(up[i], d[i]);
    ll L = inf, R = -inf;
    for(ll i = 0; i <= n; ++i) if(low[i] <= up[i]) L = min(L, low[i]), R = max(R, up[i]);
    if(L > R) printf("-1");
    for(ll i = L; i <= R; ++i) if(!((sum[n]-i)&1)) printf("%lld ",i);puts("");
    return 0;
}