【网络流24题】P4016 负载平衡问题 最小费用最大流

题目描述
GG 公司有 nn 个沿铁路运输线环形排列的仓库，每个仓库存储的货物数量不等。如何用最少搬运量可以使 nn 个仓库的库存数量相同。搬运货物时，只能在相邻的仓库之间搬运。

思路：可以唯一确定最后各自都相等时的值$each=sum/n$。将a[i]重新赋值为a[i] -= each。
建图：对于a[i]为正的部分，相当于流出量，由s连出。对于a[i]为负的部分，相当于流入量，连向t，费用都为0。然后相邻的再连边，流量为inf，费用为1（相当于每个数流动时需要消耗1点花费）。这个时候求个最小费用最大流即可。
最大流保证可以平均分配，最小费用保证答案最优。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '\n'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 2e2+200;
const ll inf=0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

const int V = 300, E = 1500;

ll head[V], pnt[E], nxt[E], cost[E], f[E], pre[E], node[E],cur[V], e = 0;

ll d[maxn];
ll vis[maxn];
ll flow[maxn];
ll a[maxn];
ll n, m;
ll s,t;
ll maxFlow = 0;
ll minCost = 0;

inline void addedge(ll u, ll v, ll ff, ll money)
{
    pnt[e] = v;
    node[e] = u;
    cost[e] = money;
    f[e] = ff;
    nxt[e] = head[u];
    head[u] = e++;
}

inline void autoAdd(ll u, ll v, ll ff, ll money)
{
    addedge(u,v,ff,money), addedge(v,u,0,-money);
}

bool SPFA()
{
    mem(d,inf);
    mem(vis,0);
    queue<ll> q;
    q.push(s);
    vis[s] = 1, d[s] = 0, pre[t] = -1;
    while(!q.empty())
    {
        ll now = q.front();
        q.pop();
        vis[now] = 0;
        for(ll j = head[now]; j != -1; j = nxt[j])
        {
            ll i = pnt[j];
            ll val = cost[j];
            ll curFlow = f[j];
            if(i!=now&&curFlow>0&&d[i]>d[now]+val)
            {
                d[i] = d[now] + val;
                if(!vis[i])
                {
                    vis[i] = 1;
                    q.push(i);
                }
            }
        }
    }
    return d[t]!=inf;
}

ll dfs(ll u, ll low)
{
    if(u==t)
    {
        vis[t] = 1;
        maxFlow += low;
        return low;
    }
    int used = 0;
    vis[u] = 1;
    for(ll i=head[u]; i!=-1; i=nxt[i])
    {
        ll v = pnt[i];
        if((!vis[v]||v==t)&&f[i]&&d[v]==d[u]+cost[i])
        {
            ll mi = dfs(v,min(low-used,f[i]));
            if(mi) minCost += cost[i]*mi, f[i] -= mi, f[i^1] += mi, used += mi;
            if(used==low) break;
        }
    }
    return used;
}


void MCMF(int s, int t)
 {
     while(SPFA())
     {
        vis[t] = 1;
        while(vis[t])
        {
            mem(vis,0);
            dfs(s,inf);
        }
     }
 }

int main()
{
    n = read();
    mem(head,-1);

    ll sum = 0;
    rep(i,1,n) a[i] = read(), sum += a[i];
    ll each = sum/n;
    rep(i,1,n) a[i] -= each;
    s = 0, t = n+1;
    rep(i,1,n)
    {
        if(a[i]>0) autoAdd(s,i,a[i],0);
        else if(a[i]<0) autoAdd(i,t,-a[i],0);
    }
    rep(i,2,n)
    {
        autoAdd(i-1,i,inf,1);
		autoAdd(i,i-1,inf,1);
    }
    autoAdd(1,n,inf,1);
    autoAdd(n,1,inf,1);
    MCMF(s,t);
    cout<<minCost<<endl;
    return 0;
}