洛谷-P4016 负载平衡问题

负载平衡问题

费用流

主要是建图，其他都是模板

• 每个仓库视为一个点，设置一个源点，对仓库有一条边，容量就为仓库原有的库存量，费用为 0

• 然后相邻的仓库之间有一条容量为无穷大，费用为 1 的边

• 每个仓库对汇点有一个边，容量为平均值，费用为 0

#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;
const int maxn = 1e3 + 10;
const int inf = 1e9 + 10;
int tp = 1, head[maxn], nex[maxn], val[maxn], to[maxn], cost[maxn];
int cur[maxn], a[maxn], s, t, n, vis[maxn], dis[maxn];
int last = 0;

void add(int u, int v, int f, int c)
{
    tp++;
    nex[tp] = head[u];
    head[u] = tp;
    cost[tp] = c;
    val[tp] = f;
    to[tp] = v;
}

bool spfa()
{
    queue<int>q;
    for(int i=0; i<=n; i++) dis[i] = inf;
    for(int i=0; i<=n; i++) cur[i] = head[i];
    dis[s] = 0;
    q.push(s);
    while(q.size())
    {
        int now = q.front();
        q.pop();
        vis[now] = 0;
        for(int i=head[now]; i; i=nex[i])
        {
            int v = to[i];
            if(val[i] > 0 && dis[now] + cost[i] < dis[v])
            {
                dis[v] = dis[now] + cost[i];
                if(vis[v] == 0)
                {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return dis[t] != inf;
}

int dfs(int now, int flow)
{
    if(now == t) return flow;
    vis[now] = 1;
    int ans = 0;
    for(int i=cur[now]; i && ans < flow; i=nex[i])
    {
        cur[now] = i;
        int v = to[i];
        if(vis[v] == 0 && val[i] > 0 && dis[v] == dis[now] + cost[i])
        {
            int x = dfs(v, min(flow - ans, val[i]));
            last += cost[i] * x;
            val[i] -= x;
            val[i ^ 1] += x;
            ans += x;
        }
    }
    vis[now] = 0;
    return ans;
}

void dinic()
{
    while(spfa())
        dfs(s, inf);
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    int sum = 0;
    for(int i=1; i<=n; i++) {cin >> a[i]; sum += a[i];}
    for(int i=1; i<=n; i++)
    {
        int x = i + 1;
        if(x > n) x = 1;
        add(i, x, sum, 1);
        add(x, i, 0, -1);
        add(x, i, sum, 1);
        add(i, x, 0, -1);
    }
    s = n + 1;
    t = s + 1;
    for(int i=1; i<=n; i++)
    {
        add(s, i, a[i], 0);
        add(i, s, 0, 0);
    }
    sum /= n;
    for(int i=1; i<=n; i++)
    {
        add(i, t, sum, 0);
        add(t, i, 0, 0);
    }
    n += 2;
    dinic();
    cout << last << endl;
    return 0;
}