版权声明：https://blog.csdn.net/qq_41730082 https://blog.csdn.net/qq_41730082/article/details/88827166
题目链接
注意题意：环形——不然怎么也算不明白为什么样例是那样的……
发现这道题竟然是一道费用流，是因为移动每一个货物可以看成是挪动一次代价为1，这样子想就可以建边了，从源点出发到每个节点是每个节点的拥有数，而每个节点到汇点的容量是那个平均数。然后对于每个相邻的节点连接上边并且单位流的代价是1。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int S = 0, maxN = 107, maxE = 1e3 + 7;
int N, head[maxN], cnt, all, T, pre[maxN], Flow[maxN], dist[maxN];
bool inque[maxN];
queue<int> Q;
struct Eddge
{
int nex, u, v, flow, cost;
Eddge(int a=-1, int b=0, int c=0, int d=0, int f=0):nex(a), u(b), v(c), flow(d), cost(f) {}
}edge[maxE];
inline void addEddge(int u, int v, int flow, int cost)
{
edge[cnt] = Eddge(head[u], u, v, flow, cost);
head[u] = cnt++;
}
inline void _add(int u, int v, int flow, int cost) { addEddge(u, v, flow, cost); addEddge(v, u, 0, -cost); }
bool spfa()
{
memset(pre, -1, sizeof(pre)); memset(dist, INF, sizeof(dist)); memset(inque, false, sizeof(inque));
while(!Q.empty()) Q.pop();
Q.push(S); inque[S] = true; Flow[S] = INF; dist[S] = 0;
while(!Q.empty())
{
int u= Q.front(); inque[u] = false; Q.pop();
for(int i=head[u], v, flow, cost; ~i; i=edge[i].nex)
{
v = edge[i].v; flow = edge[i].flow; cost = edge[i].cost;
if(flow && dist[v] > dist[u] + cost)
{
dist[v] = dist[u] + cost;
pre[v] = i;
Flow[v] = min(Flow[u], flow);
if(!inque[v])
{
inque[v] = true;
Q.push(v);
}
}
}
}
return pre[T] ^ -1;
}
int Out_Put()
{
int ans = 0;
while(spfa())
{
int now = T, last = pre[now];
while(now)
{
edge[last].flow -= Flow[T];
edge[last ^ 1].flow += Flow[T];
now = edge[last].u;
last = pre[now];
}
ans += Flow[T] * dist[T];
}
return ans;
}
inline void init()
{
cnt = all = 0; T = N + 1;
memset(head, -1, sizeof(head));
}
int main()
{
scanf("%d", &N);
init();
for(int i=1, e1; i<=N; i++)
{
scanf("%d", &e1);
all += e1;
_add(S, i, e1, 0);
if(i > 1) _add(i, i - 1, INF, 1);
else _add(i, N, INF, 1);
if(i < N) _add(i, i + 1, INF, 1);
else _add(i, 1, INF, 1);
}
all /= N;
for(int i=1; i<=N; i++) _add(i, T, all, 0);
printf("%d\n", Out_Put());
return 0;
}