「网络流 24 题」负载平衡

题目描述

G 公司有 $n$ 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 $n$ 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。

输入格式

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

输出格式

输出最少搬运量。

样例

样例输入

517 9 14 16 4

样例输出

11

数据范围与提示

$1\le n\le 100$

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

$连一容量为INF费用为的边。求一遍最小费用最大流。$

#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;}