「网络流 24 题」负载平衡

思路

首先我们从源点向每个仓库连边，容量为 $a_i$，费用为 $0$；既然所有仓库物品相同，那么数量一定是总物品的平均值，我们提前算出来 $avg$，然后从每个仓库向汇点连边，容量为 $avg$，费用为 $0$，最后在相邻仓库之间连上容量 $\infty$，费用为 $1$ 的边，跑最小费用最大流即可

#include<bits/stdc++.h>
#define fore(i,l,r)	for(int i=(int)(l);i<(int)(r);++i)
#define fi first
#define se second
#define endl '\n'
#define ull unsigned long long
#define ALL(v) v.begin(), v.end()
#define Debug(x, ed) std::cerr << #x << " = " << x << ed;const int INF=0x3f3f3f3f;
const long long INFLL=1e18;typedef long long ll;struct MCF {struct Edge {int v, c, w; //边终点、容量、费用Edge(int v, int c, int w) : v(v), c(c), w(w) {}};const int n;std::vector<Edge> e;std::vector<std::vector<int>> g;std::vector<ll> h, dis;std::vector<int> pre;bool dijkstra(int s, int t) {dis.assign(n + 1, std::numeric_limits<ll>::max());pre.assign(n + 1, -1);std::priority_queue<std::pair<ll, int>, std::vector<std::pair<ll, int>>, std::greater<std::pair<ll, int>>> que;dis[s] = 0;que.emplace(0, s);while (!que.empty()) {ll d = que.top().first;int u = que.top().second;que.pop();if (dis[u] < d) continue;for (int i : g[u]) {int v = e[i].v;int c = e[i].c;int w = e[i].w;if (c > 0 && dis[v] > d + h[u] - h[v] + w) {dis[v] = d + h[u] - h[v] + w;pre[v] = i;que.emplace(dis[v], v);}}}return dis[t] != std::numeric_limits<ll>::max();}MCF(int n) : n(n), g(n + 1) {}void addEdge(int u, int v, int c, int w) {g[u].push_back(e.size());e.emplace_back(v, c, w);g[v].push_back(e.size());e.emplace_back(u, 0, -w);}std::pair<int, ll> flow(int s, int t) {int flow = 0;ll cost = 0;h.assign(n + 1, 0);while (dijkstra(s, t)) {for (int i = 1; i <= n; ++i) h[i] += dis[i];int aug = std::numeric_limits<int>::max();for (int i = t; i != s; i = e[pre[i] ^ 1].v) aug = std::min(aug, e[pre[i]].c);for (int i = t; i != s; i = e[pre[i] ^ 1].v) {e[pre[i]].c -= aug;e[pre[i] ^ 1].c += aug;}flow += aug;cost += ll(aug) * h[t];}return std::make_pair(flow, cost);}
};int main(){std::ios::sync_with_stdio(false);std::cin.tie(nullptr);std::cout.tie(nullptr);int n;std::cin >> n;MCF mcf(n + 2);int S = n + 1, T =  S + 1;std::vector<int> a(n + 1, 0);fore(i, 1, n + 1){std::cin >> a[i];mcf.addEdge(S, i, a[i], 0);}int avg = std::accumulate(ALL(a), 0) / n;fore(i, 1, n + 1) mcf.addEdge(i, T, avg, 0);fore(i, 2, n + 1){mcf.addEdge(i - 1, i, INF, 1);mcf.addEdge(i, i - 1, INF, 1);}mcf.addEdge(1, n, INF, 1);mcf.addEdge(n, 1, INF, 1);std::cout << mcf.flow(S, T).se;return 0;
}