洛谷P4009骑车加油行驶问题（分层图、最小费用最大流） 洛谷P4009骑车加油行驶问题

题目链接

建一个(k+1)层的图，第0层为到加油站，第(i) ((1leq i leq k))层为走了(i)步，每个点向下一层中与它四联通的点建流量为1花费为0的边。需要注意它只要经过加油站就必须要加油，所以在1到k层中，加油站的点只能向第0层建边，不可向下一层建边

#include <bits/stdc++.h>

using namespace std;
const int maxn = 300100;
const int maxm = 3000100;
const int inf = 0x3f3f3f3f;
typedef long long ll;
struct edge {
    int to, next, cap, flow, cost;
} e[maxm];
int head[maxn], tot;
int pre[maxn], dis[maxn];
bool vis[maxn];
int N;

void init(int n) {
    N = n;
    tot = 1;
    memset(head, 0, sizeof(head));
}

void addedge(int u, int v, int cap, int cost) {
    e[++tot].to = v, e[tot].next = head[u], e[tot].cap = cap, e[tot].cost = cost, e[tot].flow = 0, head[u] = tot;
    e[++tot].to = u, e[tot].next = head[v], e[tot].cap = 0, e[tot].flow = 0, e[tot].cost = -cost, head[v] = tot;
}

bool spfa(int s, int t) {
    deque<int> q;
    for (int i = 0; i <= N; ++i) {
        dis[i] = inf;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push_front(s);
    while (!q.empty()) {
        int u = q.front();
        q.pop_front();
        vis[u] = 0;
        for (int i = head[u]; i; i = e[i].next) {
            int v = e[i].to;
            if (e[i].cap > e[i].flow && dis[v] > dis[u] + e[i].cost) {
                dis[v] = dis[u] + e[i].cost;
                pre[v] = i;
                if (!vis[v]) {
                    vis[v] = true;
                    if (!q.empty()) {
                        if (dis[v] < dis[q.front()]) q.push_front(v);
                        else q.push_back(v);
                    } else q.push_back(v);
                }
            }
        }
    }
    return pre[t] != -1;
}

//return 最大流，cost为最小费用
int mcfc(int s, int t, ll &cost) {
    cost = 0;
    int flow = 0;
    while (spfa(s, t)) {
        int minn = inf;
        for (int i = pre[t]; i != -1; i = pre[e[i ^ 1].to]) {
            if (minn > e[i].cap - e[i].flow) {
                minn = e[i].cap - e[i].flow;
            }
        }
        for (int i = pre[t]; i != -1; i = pre[e[i ^ 1].to]) {
            e[i].flow += minn;
            e[i ^ 1].flow -= minn;
            cost += 1ll * e[i].cost * minn;
        }
        flow += minn;
    }
    return flow;
}

int mp[200][200];
int n;

inline int getid(int deep, int i, int j) {
    return deep * n * n + (i - 1) * n + j;
}

int main() {
//    freopen("in.txt", "r", stdin);
    int k, A, B, C;
    scanf("%d %d %d %d %d", &n, &k, &A, &B, &C);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            scanf("%d", &mp[i][j]);
        }
    }
    int s = 0, t = getid(k + 1, 1, 1);
    init(t + 1);
    addedge(s, getid(0, 1, 1), 1, 0);
    for (int deep = 0; deep <= k; deep++) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if (mp[i][j] && deep) {
                    addedge(getid(deep, i, j), getid(0, i, j), 1, A);
                    continue;
                }
                if (deep < k) {
                    if (j + 1 <= n)
                        addedge(getid(deep, i, j), getid(deep + 1, i, j + 1), 1, 0);
                    if (i + 1 <= n)
                        addedge(getid(deep, i, j), getid(deep + 1, i + 1, j), 1, 0);
                    if (j - 1 >= 1)
                        addedge(getid(deep, i, j), getid(deep + 1, i, j - 1), 1, B);
                    if (i - 1 >= 1)
                        addedge(getid(deep, i, j), getid(deep + 1, i - 1, j), 1, B);
                }
                int cost;
                if (deep == 0) continue;
                if (mp[i][j])cost = A;
                else cost = C + A;
                addedge(getid(deep, i, j), getid(0, i, j), 1, cost);
            }
        }
        addedge(getid(deep, n, n), t, 1, 0);
    }
    ll ans = 0;
    mcfc(s, t, ans);
    printf("%lld
", ans);
    return 0;
}