网络流二十四题之负载问题

这道题其实是拆点费用流，一开始以为判断下残量网络但是不行：
1.源点向每个在X集合的点，连一条边，容量为ai，费油为0，Y集合点向汇点连一条边，容量是平均值，费用为0；
2.X集合向对应的在Y集合的点连一条边，容量INF，费用为0
3.按照环的顺序从Y集合到X集合连边，容量INF，费用为1，这样做的原因是因为转移的过程不可能只进行一次，有可能对于一个点发生了多次转移和接受
之后跑一次费用流就可以了

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to,next,cap,flow,cost;
} edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
    N = n;
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa(int s,int t)
{
    queue<int>q;
    for(int i = 0; i < N; i++)
    {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(edge[i].cap > edge[i].flow &&
                    dis[v] > dis[u] + edge[i].cost )
            {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t] == -1)
        return false;
    else
        return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost)
{
    int flow = 0;
    cost = 0;
    while(spfa(s,t))
    {
        int Min = INF;
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            if(Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}
int a[1001];

int main()
{
    int n;
    scanf("%d",&n);
    int sum = 0;
    for(int i = 1; i<=n; i++)
    {
        scanf("%d",&a[i]);
        sum += a[i];
    }
    sum /= n;
    init(2*n + 2);
    for(int i = 1; i<=n; i++)
    {
        addedge(0,i,a[i],0);
        addedge(i + n, 2*n + 1,sum,0);
        addedge(i,i+n,INF,0);
        if(i != 1)
            addedge(i+n,i-1,INF,1);
        if(i != n)
            addedge(i+n,i+1,INF,1);
    }
    addedge(n +n,1,INF,1);
    addedge(n +1,n,INF,1);
    int cost = 0;
    int ans = minCostMaxflow(0,2*n+1,cost);
    printf("%d\n",cost);
    return 0;
}