Codeforces 543B Destroying Roads(最短路)

#include <bits/stdc++.h>
using namespace std;

const int N = 3009;

struct edge {
    int v, ne;
} E[N * N << 1];
int head[N], cnt;

int n, m, ans;
int s1, s2, t1, t2, l1, l2;

int dis[5][N];
int vis[N];

void SPFA (int S, int k) {
    queue<int> q;
    memset (vis, 0, sizeof vis);
    dis[k][S] = 0;
    vis[S] = 1;
    q.push (S);
    while (!q.empty() ) {
        int u = q.front(); q.pop();
        for (int i = head[u]; i; i = E[i].ne) {
            int v = E[i].v;
            if (dis[k][u] + 1 < dis[k][v]) {
                dis[k][v] = dis[k][u] + 1;
                if (!vis[v]) {
                    vis[v] = 1;
                    q.push (v);
                }
            }
        }
        vis[u] = 0;
    }
}

inline void  add (int u, int v) {
    E[++cnt].v = v, E[cnt].ne = head[u];
    head[u] = cnt;
}

int main() {
    ios::sync_with_stdio (0);
    cin >> n >> m;
    for (int i = 1, u, v; i <= m; ++i) {
        cin >> u >> v;
        add (u, v), add (v, u);
    }
    cin >> s1 >> t1 >> l1;
    cin >> s2 >> t2 >> l2;
    memset (dis, 0x1f, sizeof dis);
    SPFA (s1, 0);
    SPFA (s2, 1);
    SPFA (t1, 2);
    SPFA (t2, 3);
    if (dis[0][t1] <= l1 && dis[1][t2] <= l2) {
        ans = dis[0][t1] + dis[1][t2];
        for (int i = 1; i <= n; ++i) {
            int flag = 0;
            for (int k = 1; k <= n; ++k) dis[4][k] = 33333;
            for (int j = i + 1; j <= n; ++j) {
                int tem1 = min (dis[0][i] + dis[2][j], dis[0][j] + dis[2][i]);
                int tem2 = min (dis[1][i] + dis[3][j], dis[1][j] + dis[3][i]);
                if (tem1 + tem2 < ans && tem1 < l1 && tem2 < l2) {
                    if (!flag) {
                        SPFA (i, 4);
                        flag = 1;
                    }
                    if (tem1 + dis[4][j] <= l1 && tem2 + dis[4][j] <= l2) {
                        ans = min (ans, tem1 + tem2 + dis[4][j]);
                    }
                }
            }
        }
    }
    else {
        ans = m + 1;
    }
    cout << m - ans << endl;
}