P2053 [SCOI2007]修车 费用流

https://www.luogu.org/problemnew/show/P2053

题意

同一时刻有N位车主带着他们的爱车来到了汽车维修中心。维修中心共有M位技术人员，不同的技术人员对不同的车进行维修所用的时间是不同的。现在需要安排这M位技术人员所维修的车及顺序，使得顾客平均等待的时间最小。

说明：顾客的等待时间是指从他把车送至维修中心到维修完毕所用的时间。

思路

左边放n个点，表示n辆车。右边放m * n个点，表示m个工人，拆除n个点表示不同阶段。即如果第i辆车连上了第m个工人的第k个阶段，表示第m个工人第k阶段才修理第i辆车。
假设第i辆车的等待时间是wi

那么总等待时间是$n*w1 + (n-1)*w2 + …… + 1*wn$;

所以越早修理，需要的时间是多倍的。
$addedge(i, n + (j-1)*n + k, 1, mp[i][j] * (n - k + 1))$;

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/
const int maxn = 609;

struct E {
    int v, val;
    int cost;
    int nxt;
} edge[80009];
int head[maxn], gtot = 0;
void addedge(int u, int v, int val, int cost) {
    edge[gtot].v = v;
    edge[gtot].val = val;
    edge[gtot].cost = cost;
    edge[gtot].nxt = head[u];
    head[u] = gtot++;

    edge[gtot].v = u;
    edge[gtot].val = 0;
    edge[gtot].cost = -cost;
    edge[gtot].nxt = head[v];
    head[v] = gtot++;
}
int mp[maxn][maxn];

int dis[maxn], path[maxn], vis[maxn], pre[maxn];
bool spfa(int s, int t) {
    memset(pre, -1, sizeof(pre));
    memset(dis, inf, sizeof(dis));
    memset(vis, 0, sizeof(vis));

    queue<int> que;
    que.push(s);
    vis[s] = 1;
    dis[s] = 0;
    while (!que.empty()) {
        int u = que.front();
        que.pop();
        vis[u] = 0;
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            int v = edge[i].v, val = edge[i].val, cost = edge[i].cost;

            if (val > 0 && dis[v] > dis[u] + cost) {
                dis[v] = dis[u] + cost;
                pre[v] = u;
                path[v] = i;
                if (vis[v] == 0) {
                    vis[v] = 1;
                    que.push(v);
                }
            }
        }
    }
    return pre[t] != -1;
}
int mcmf(int s, int t) {
    int flow = 0, cost = 0;
    while (spfa(s, t)) {
        int f = inf;
        for (int i = t; i != s; i = pre[i]) {
            f = min(f, edge[path[i]].val);
        }
        flow += f;
        cost += f * dis[t];
        for (int i = t; i != s; i = pre[i]) {
            edge[path[i]].val -= f;
            edge[path[i] ^ 1].val += f;
        }
    }
    return cost;
}
int main() {
    memset(head, -1, sizeof(head));
    int n, m;
    scanf("%d%d", &m, &n);
    int s = 0, t = n + m * n + 1;
    rep(i, 1, n) {
        addedge(s, i, 1, 0);
        rep(j, 1, m) {
            scanf("%d", &mp[i][j]);
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            for (int t = 1; t <= n; t++) {
                addedge(i, n + (j - 1) * n + t, 1, mp[i][j] * (n - t + 1));
            }
        }
    }
    for (int i = n + 1; i < t; i++) addedge(i, t, 1, 0);
    printf("%.2f\n", 1.0 * mcmf(s, t) / n);
    return 0;
}