Codechef July Challenge 2019 Snake and Apple Tree

费用流。
把每个方格拆成 $T$ 个点，$t$ 时刻一个方格向周围四个方格的 $t + 1$ 的点连一条容量为 $1$ 费用为 $0$ 的边，向自身的 $t + 1$ 连一条容量为 $1$ 费用为该方格最大幸福值的边。
源点向方格为 'S' 的0时刻连一条容量为 $1$ 费用为 $0$ 的边。所有点的 $T - 1$ 时刻向汇点连一条容量为 $1$ 费用为该方格最大幸福值的边。
还有每个格子同时刻不能有多条蛇呆在上面，再把每个点每个时刻拆成两个点，容量为 $1$ 费用为 $0$。跑最大费用最大流即可。

#include <bits/stdc++.h>
using namespace std;
 
const int N = ;
const int NN = 1e5 + , M = 5e5 + ;
const int INF = 0x3f3f3f3f;
 
template<class T>
inline void checkmax(T &a, T b) {
    if (a < b) a = b;
}
template<class T>
inline void checkmin(T &a, T b) {
    if (a > b) a = b;
}
 
struct E {
    int v, ne, f, c;
} e[M];
int head[NN], cnt, tol;
int id[N][N][], mp[N][N][];
int dis[NN], path[NN], n, m, z, t;
bool inq[NN];
char str[N][N];
const int dx[] = {, , , -}, dy[] = {-, , , };
 
inline void add(int u, int v, int f, int c) {
    e[cnt].v = v; e[cnt].f = f; e[cnt].c = c; e[cnt].ne = head[u]; head[u] = cnt++;
    e[cnt].v = u; e[cnt].f = ; e[cnt].c = -c; e[cnt].ne = head[v]; head[v] = cnt++;
}
 
bool spfa(int s, int t) {
    for (int i = ; i <= t; i++)
        dis[i] = INF, inq[i] = , path[i] = -;
    dis[s] = ;
    inq[s] = ;
    queue<int> que;
    que.push(s);
    while (!que.empty()) {
        int u = que.front(); que.pop();
        inq[u] = ;
        for (int i = head[u]; ~i; i = e[i].ne) {
            int v = e[i].v, c = e[i].c;
            if (e[i].f && dis[v] > dis[u] + c) {
                dis[v] = dis[u] + c;
                path[v] = i;
                if (!inq[v]) {
                    inq[v] = ;
                    que.push(v);
                }
            }
        }
    }
    return dis[t] != INF;
}
 
int mcf(int s, int t) {
    int ans = ;
    while (spfa(s, t)) {
        for (int i = path[t]; ~i; i = path[e[i ^ ].v]) e[i].f--, e[i ^ ].f++;
        ans += dis[t];
    }
    return ans;
}
 
int main() {
    //freopen("in.txt", "r", stdin);
    memset(head, -, sizeof(head));
    scanf("%d%d%d%d", &n, &m, &z, &t);
    for (int i = ; i <= n; i++)
        scanf("%s", str[i] + );
    for (int i = ; i <= z; i++) {
        int x, y, p, q, h;
        scanf("%d%d%d%d%d", &x, &y, &p, &q, &h);
        for (int j = p; j < q; j++)
            checkmax(mp[x][y][j], h);
    }
    tol = ;
    for (int i = ; i <= n; i++)
        for (int j = ; j <= m; j++) if (str[i][j] != '#')
            for (int k = ; k < t; k++) {
                add(tol, tol + , , );
                //cout << tol << ' ' << tol + 1 << endl;
                id[i][j][k] = tol;
                tol += ;
            }
    int S = tol + , T = tol + ;
    for (int i = ; i <= n; i++)
        for (int j = ; j <= m; j++) if (id[i][j][]) {
            for (int d = ; d < ; d++) if (id[i + dx[d]][j + dy[d]][])
                for (int k = ; k < t - ; k++)
                    add(id[i][j][k] + , id[i + dx[d]][j + dy[d]][k + ], , );
            for (int k = ; k < t - ; k++)
                add(id[i][j][k] + , id[i][j][k + ], , -mp[i][j][k]);
            add(id[i][j][t - ] + , T, , -mp[i][j][t - ]);
            if (str[i][j] == 'S') add(S, id[i][j][], , );
        }
    printf("%d\n", -mcf(S, T));
    return ;
}