Codeforces 1091E

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題意：給定n個點的度數，請你添加第n+1個點，輸出所有可能的第n+1個點的度數

做法：1. 查看鏈接知道了下面這個定理

A sequence of non-negative integers $ d_1\geq\cdots\geq 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 \({\displaystyle 1\leq k\leq n}\).

2.排序之后，\([low(p), up(p)]\) 表示在\([p, p+1]\) 之間插入新的點度數\(d_t\)的取值范圍。

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

(2) 當 \(d_t\) 在不等式左側時
\[ d_t \leq k(k+1)+\sum _{i=k+1}^{n}\min(d_{i},k+1) - \sum _{i=1}^{k}d_{i}, p \leq k \leq n \\ \Downarrow \\ up(p) = min(up(p + 1) , p(p+1) + \sum_{i=p+1}^n min(d_i,p+1) - \sum_{i=1}^p d_i) \]

(3) 當 \(d_t\) 在不等式右側時
\[ \sum_{i=1}^k d_i - k(k-1) - \sum_{i=k+1}^n min(d_i,k) <= d_t , 1 \leq k \leq p \\ \Downarrow\\ low(p) = max(\sum_{i=1}^p d_i - p(p-1) - \sum_{i=p+1}^n min(d_i,p), low(p-1)) \\ \]

3.猜測答案為連續的一段值，將上一步求出的區間并起來

4.又因為無向簡單圖，每條邊對度數貢獻為2，由此可以確定答案的奇偶性

5.確定上下界時，需要維護\(\sum_{i = k+1}^{n} min(d_i,k)\) 和\(\sum_{i = k}^{n} min(d_i,k)\) 我們倒著枚舉\(k\)，用優先隊列維護小于等于當前\(k\)的集合，以及這個集合內所有元素的和，就可以預處理出這個式子，因為k是遞減的因此保證了正確性

6.一定要分析清楚再寫。。。代碼細節很多

#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; }

轉載于:https://www.cnblogs.com/RRRR-wys/p/10203438.html

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