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bzoj4873 [Shoi2017]寿司餐厅
传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=4873
【题解】
没看出来是最大权闭合子图模型……要多学习学习qwq
首先区间$[i,j]$依赖于区间$[i+1,j]$和$[i,j-1]$。每个区间$[i,j](i < j)$的权值就是$d_{i,j}$。
特别地,区间$[i, i]$的权值为$d_{i,j} - a_i$(由于花费原因)。
区间$[i, i]$还依赖于$a_i$这个点,$a_i$这个点的权值为$-m * a_i * a_i$(花费)。
然后直接跑最大权闭合子图模型即可。
时间复杂度O(能过)。
这个区间建模有点意思,看来知识水平不够高啊qwq
# include <queue> # include <stdio.h> # include <string.h> # include <iostream> # include <algorithm> // # include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; const int M = 8e5 + 10, N = 1e2 + 10; const int mod = 1e9+7; const ll inf = 1e16; int n, K, idx, S, T; ll sum = 0; vector<int> ps; int a[N], d[N][N], id[N][N], bid[N * 10]; namespace MF { int head[M], nxt[M], to[M], tot = 1; ll flow[M]; inline void add(int u, int v, ll fl) { ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v; flow[tot] = fl; } inline void adde(int u, int v, ll fl) { add(u, v, fl), add(v, u, 0ll); } queue<int> q; int c[M], cur[M]; inline bool bfs() { while(!q.empty()) q.pop(); for (int i=1; i<=idx; ++i) c[i] = -1; c[S] = 0; q.push(S); while(!q.empty()) { int top = q.front(); q.pop(); for (int i=head[top]; i; i=nxt[i]) { if(c[to[i]] != -1 || !flow[i]) continue; c[to[i]] = c[top] + 1; q.push(to[i]); if(to[i] == T) return 1; } } return 0; } inline ll dfs(int x, ll low) { if(x == T) return low; ll r = low, fl; for (int i=cur[x]; i; i=nxt[i]) { if(c[to[i]] != c[x] + 1 || !flow[i]) continue; fl = dfs(to[i], min(r, flow[i])); flow[i] -= fl; r -= fl; flow[i^1] += fl; if(flow[i] > 0) cur[x] = i; if(!r) return low; } if(low == r) c[x] = -1; return low-r; } inline ll main() { ll ret = 0; while(bfs()) { for (int i=1; i<=idx; ++i) cur[i] = head[i]; ret += dfs(S, inf); } return ret; } } inline void add(int u, ll d) { if(d >= 0) MF :: adde(S, u, d), sum += d; else MF :: adde(u, T, -d); } int main() { cin >> n >> K; for (int i=1; i<=n; ++i) { scanf("%d", &a[i]); ps.push_back(a[i]); } S = ++idx, T = ++idx; for (int i=1; i<=n; ++i) for (int j=i; j<=n; ++j) { scanf("%d", &d[i][j]); id[i][j] = ++idx; if(i == j) d[i][j] -= a[i]; } for (int i=1; i<=n; ++i) for (int j=i; j<=n; ++j) add(id[i][j], d[i][j]); sort(ps.begin(), ps.end()); ps.erase(unique(ps.begin(), ps.end()), ps.end()); for (int i=0; i<ps.size(); ++i) { bid[ps[i]] = ++idx; add(bid[ps[i]], -1ll * K * ps[i] * ps[i]); } for (int i=1; i<=n; ++i) { for (int j=i+1; j<=n; ++j) { MF :: adde(id[i][j], id[i+1][j], inf); MF :: adde(id[i][j], id[i][j-1], inf); } MF :: adde(id[i][i], bid[a[i]], inf); } cout << max(sum - MF :: main(), 0ll); return 0; }
bzoj4873 [Shoi2017]寿司餐厅
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