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P4016 负载平衡问题

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P4016 负载平衡问题

题面

解法

显然有贪心解法，但为了练习费用流选择使用网络流解法。

第一次写$spfa$和费用流，没想到会出现这么多问题。。。

有几个要注意的点：

  1. 费用流建反边的$cost$是原来的$cost$的相反数，即需要add(u, v, c, f), add(v, u, -c, 0);
  2. 计算流量的方式和普通网络流相同，但是计算总$cost$是每次加上edge.cost * flow，不要忘记乘上$cost$，用全局变量存$cost$会好写一些。
  3. 由于费用流的边权可能$\le 0$，所以必须用$spfa$而无法使用$dijkstra$，并且$dfs$增广时必须开$vis$数组防止无限递归。
  4. （这应该算是个小$tip$）网络流$tot$一开始清空成$1$可以不需要再调用init函数，也不需要再把$head$清空成$-1$。

AC代码

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/**
 * @file:           P4016.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P4016
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
    bool sign = false;
    char ch = getchar();
    long double tmp = 1;
    for (; !isdigit(ch); ch = getchar())
        sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar())
        x = x * 10 + (ch ^ 48);
    if (ch == '.')
        for (ch = getchar(); isdigit(ch); ch = getchar())
            tmp /= 10.0, x += tmp * (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        write(x / 10);
    putchar((x % 10) ^ 48);
}
const int MAXN = 150;
const int MAXM = 1e3 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct Dinic {
    struct Edge {
        int v, flow, cost;
    } edge[MAXM];
    int tot = 1, flow = 0, cost = 0;
    int head[MAXN], nxt[MAXM], dis[MAXN], cur[MAXN];
    bool vis[MAXN];
    void addedge(int u, int v, int flow, int cost) {
        edge[++tot] = {v, flow, cost};
        nxt[tot] = head[u], head[u] = tot;
        edge[++tot] = {u, 0, -cost};
        nxt[tot] = head[v], head[v] = tot;
    }
    bool spfa(int s, int t) {
        fill(vis, vis + MAXN, 0);
        fill(dis, dis + MAXN, INF);
        queue<int> que;
        que.push(s);
        dis[s] = 0, vis[s] = 1;
        while (!que.empty()) {
            int u = que.front();
            que.pop(), vis[u] = 0;
            for (int i = head[u]; i; i = nxt[i]) {
                int v = edge[i].v;
                if (edge[i].flow && dis[v] > dis[u] + edge[i].cost) {
                    dis[v] = dis[u] + edge[i].cost;
                    if (!vis[v])
                        que.push(v), vis[v] = 1;
                }
            }
        }
        return dis[t] != INF;
    }
    int augment(int u, int t, int mx) {
        if (u == t || mx == 0)
            return mx;
        vis[u] = 1;
        int ret = 0;
        for (int &i = cur[u]; i; i = nxt[i]) {
            int v = edge[i].v;
            if (vis[v] || dis[v] != dis[u] + edge[i].cost)
                continue;
            int tmp = augment(v, t, min(mx, edge[i].flow));
            cost += tmp * edge[i].cost;
            mx -= tmp, ret += tmp;
            edge[i].flow -= tmp, edge[i ^ 1].flow += tmp;
            if (mx == 0)
                break;
        }
        vis[u] = 0;
        return ret;
    }
    int mcmf(int s, int t) {
        while (spfa(s, t)) {
            copy(head, head + MAXN, cur);
            flow += augment(s, t, INF);
        }
        return flow;
    }
};
int n, a[MAXN];
Dinic nf;
signed main() {
    read(n);
    int num = 0;
    for (int i = 1; i <= n; ++i)
        read(a[i]), num += a[i];
    num /= n;
    int s = n + 1, t = n + 2;
    for (int i = 1; i <= n; ++i) {
        if (a[i] > num)
            nf.addedge(s, i, a[i] - num, 0);
        if (a[i] < num)
            nf.addedge(i, t, num - a[i], 0);
        nf.addedge(i, i % n + 1, INF, 1);
        nf.addedge(i % n + 1, i, INF, 1);
    }
    nf.mcmf(s, t);
    write(nf.cost), putchar('\n');
    return 0;
}