汽车加油行驶问题

分层图，k只有10，每个k一层图
跑费用流，但容量为1，所以就是SPFA

    # include <bits/stdc++.h>
# define RG register
# define IL inline
# define ID(a, b, c) ((c) * N * N + (a - 1) * N + b)
# define Fill(a, b) memset(a, b, sizeof(a))
# define Copy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2e5 + 10), __(3e7 + 10), INF(2e9);

IL ll Read(){
    char c = '%'; ll x = 0, z = 1;
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
    return x * z;
}

int N, K, A, B, C;
int cnt, fst[_], w[__], to[__], nxt[__], dis[_], vis[_], S, T, cost[__], pe[_], pv[_], max_flow, min_cost;
queue <int> Q;

IL void Add(RG int u, RG int v, RG int f, RG int co){
    cost[cnt] = co; w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;
    cost[cnt] = -co; w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;
}

IL bool Bfs(){
    Q.push(S); Fill(dis, 127); dis[S] = 0; vis[S] = 1;
    while(!Q.empty()){
        RG int u = Q.front(); Q.pop();
        for(RG int e = fst[u]; e != -1; e = nxt[e]){
            if(!w[e] || dis[to[e]] <= dis[u] + cost[e]) continue;
            dis[to[e]] = dis[u] + cost[e];
            pe[to[e]] = e; pv[to[e]] = u;
            if(!vis[to[e]]) vis[to[e]] = 1, Q.push(to[e]);
        }
        vis[u] = 0;
    }
    if(dis[T] >= dis[T + 1]) return 0;
    RG int ret = INF;
    for(RG int u = T; u != S; u = pv[u]) ret = min(ret, w[pe[u]]);
    for(RG int u = T; u != S; u = pv[u]) w[pe[u]] -= ret, w[pe[u] ^ 1] += ret;
    min_cost += ret * dis[T]; max_flow += ret;
    return 1;
}

int main(RG int argc, RG char *argv[]){
    Fill(fst, -1);
    N = Read(); K = Read(); A = Read(); B = Read(); C = Read();
    for(RG int i = 1; i <= N; ++i)
        for(RG int j = 1; j <= N; ++j){
            RG int gg = Read();
            if(gg){
                for(RG int k = 0; k < K; ++k){
                    if(i < N) Add(ID(i, j, k), ID(i + 1, j, K - 1), 1, A);
                    if(j < N) Add(ID(i, j, k), ID(i, j + 1, K - 1), 1, A);
                    if(i > 1) Add(ID(i, j, k), ID(i - 1, j, K - 1), 1, A + B);
                    if(j > 1) Add(ID(i, j, k), ID(i, j - 1, K - 1), 1, A + B);
                }
            }
            else{
                for(RG int k = 0; k < K; ++k) Add(ID(i, j, k), ID(i, j, K), 1, A + C);
                for(RG int k = 1; k <= K; ++k){
                    if(i < N) Add(ID(i, j, k), ID(i + 1, j, k - 1), 1, 0);
                    if(j < N) Add(ID(i, j, k), ID(i, j + 1, k - 1), 1, 0);
                    if(i > 1) Add(ID(i, j, k), ID(i - 1, j, k - 1), 1, B);
                    if(j > 1) Add(ID(i, j, k), ID(i, j - 1, k - 1), 1, B);
                }
            }
        }
    T = N * N * (K + 1) + 1; Add(S, ID(1, 1, K), 1, 0);
    for(RG int i = 0; i <= K; ++i) Add(ID(N, N, i), T, 1, 0);
    while(Bfs()); printf("%d
", min_cost);
    return 0;
}