P2050 [NOI2012]美食节 动态连边优化费用流

题意

类似的一道排队等候，算最小总等待时间的题目。

思路

但是这道题的边数很多，直接跑会tle，可以动态加边，就是先连上倒数第一次操作的边，跑一遍费用流，然后对使用了倒数第一条边的点，连上相应的倒数第二条边。以此类推

 

#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 ll mod = 2147483648;
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 = 309;
int n, m;
int p[maxn], mp[maxn][maxn];

struct E {
    int v, val, cost;
    int nxt;
} edge[8000009];
int head[maxn * maxn], gtot;
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 vis[maxn * maxn], pre[maxn * maxn], path[maxn * maxn];
ll dis[maxn * maxn];
bool spfa(int s, int t) {
    for (int i = s; i <= t; i++) dis[i] = inff, pre[i] = -1, vis[i] = 0;

    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 sp = 0;
ll mcmf(int s, int t) {
    ll 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 += 1ll * f * dis[t];
        for (int i = t; i != s; i = pre[i]) {
            edge[path[i]].val -= f;
            edge[path[i] ^ 1].val += f;
        }

        int la = edge[path[t] ^ 1].v + 1;

        int p = (la - 1 - n) / sp + 1;
        int b = (la - 1 - n) % sp + 1;

        addedge(la, t, 1, 0);
        for (int i = 1; i <= n; i++) {
            addedge(i, la, 1, mp[i][p] * b);
        }
    }
    return cost;
}
int main() {
    memset(head, -1, sizeof(head));
    scanf("%d%d", &n, &m);
    rep(i, 1, n) scanf("%d", &p[i]), sp += p[i];
    rep(i, 1, n) rep(j, 1, m) scanf("%d", &mp[i][j]);
    int s = 0, t = n + m * sp + 1;

    for (int i = 1; i <= n; i++) addedge(s, i, p[i], 0);
    for (int i = 1; i <= m; i++) {
        addedge(n + (i - 1) * sp + 1, t, 1, 0);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            addedge(i, n + (j - 1) * sp + 1, 1, mp[i][j]);
        }
    }

    printf("%lld\n", mcmf(s, t));
    return 0;
}