G 公司有 n 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 n 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。

文件的第 1 11 行中有 1 11 个正整数 n nn，表示有 n nn 个仓库。
第 2 22 行中有 n nn 个正整数，表示 n nn 个仓库的库存量。

输出最少搬运量。

样例输入

5
17 9 14 16 4

样例输出

11

1≤n≤100 1 \leq n \leq 1001≤n≤100

源点连所有仓库容量为初始货量费用为0，所有仓库联想汇点容量为平均货量费用为0。所有仓库都与他相邻仓库

连一容量为INF费用为的边。求一遍最小费用最大流。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
const int maxm = 10005;
const int maxn = 100005;
const int INF = 1e9 + 7;
struct node
{int u, v, flow, cost, next;
}edge[maxn];
int dis[maxm], head[maxm], cur[maxm], pre[maxn], f[1005][1005], map[1005][1005];
int s, t, n, m, cnt;
void init()
{cnt = 0, s = 0, t = n + 1;memset(head, -1, sizeof(head));
}
void add(int u, int v, int w, int cost)
{edge[cnt].u = u, edge[cnt].v = v;edge[cnt].flow = w, edge[cnt].cost = cost;edge[cnt].next = head[u], head[u] = cnt++;edge[cnt].u = v, edge[cnt].v = u;edge[cnt].flow = 0, edge[cnt].cost = -cost;edge[cnt].next = head[v], head[v] = cnt++;
}
int bfs()
{queue<int>q;for (int i = 0;i <= 10004;i++) dis[i] = INF;memset(pre, -1, sizeof(pre));dis[s] = 0;q.push(s);int rev = 0;while (!q.empty()){int u = q.front();q.pop();for (int i = head[u];i != -1;i = edge[i].next){int v = edge[i].v;if (dis[v] > dis[u] + edge[i].cost&&edge[i].flow){dis[v] = dis[u] + edge[i].cost;pre[v] = i;q.push(v);}}}if (dis[t] == INF) return 0;return 1;
}
int MCMF()
{int minflow, ans = 0;while (bfs()){minflow = INF;for (int i = pre[t];i != -1;i = pre[edge[i].u])minflow = min(minflow, edge[i].flow);for (int i = pre[t];i != -1;i = pre[edge[i].u]){edge[i].flow -= minflow;edge[i ^ 1].flow += minflow;}ans += minflow*dis[t];}return ans;
}
int main()
{int i, j, k, sum = 0, id = 0, x;scanf("%d", &n);init();for (i = 1;i <= n;i++){scanf("%d", &k);add(s, i, k, 0);sum += k;}x = sum / n; for (i = 1;i <= n;i++){add(i, t, x, 0);add(i, i%n + 1, INF, 1);add(i%n + 1, i, INF, 1);}printf("%d\n", MCMF());return 0;
}