[codeforces-543B]bfs求最短路

题意：给一个边长为1的无向图，求删去最多的边使得从a到b距离<=f,从c到d距离<=g，a,b,c,d,f,g都是给定的，求最多删去的边数。

思路：反过来思考，用最少的边构造两条从a到b，从c到d的路径，使得它们满足题目中的条件。于是可以把这两条路径的相对位置分为两种情况，不相交和相交，对于不相交的情况答案就是两组距离之和，对于相交的情况，两条路径肯定共享一段连续路径，对于共享多段的情况可以把中间没共享的取较小值变成共享，得到的距离和不会更大，因此最优情况一定是一段连续的路径。于是可以枚举共享段的两个端点来更新答案，两点之间距离可以通过n次bfs求出。有个小地方需要注意，共享段对于两条路径来说是有方向的，所以需要正反更新。

 #pragma comment(linker, "/STACK:10240000,10240000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #define constructInt5(name, a, b, c, d, e) name(int a = 0, int b = 0, int c = 0, int d = 0, int e = 0): a(a), b(b), c(c), d(d), e(e) {}
 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << a << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 #define mp(a, b) make_pair(a, b)
 #define pb(a) push_back(a)

 typedef long long LL;
 typedef pair<int, int> pii;
 typedef vector<int> vi;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 3e3 + ;
 const int md = ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }

 struct Graph {
     vector<vector<int> > G;
     void clear() { G.clear(); }
     void resize(int n) { G.resize(n + ); }
     void add(int u, int v) { G[u].push_back(v); }
     vector<int> & operator [] (int u) { return G[u]; }
 };
 Graph G;

 int n, m, s1, s2, t1, t2, l1, l2, ans, dist[maxn][maxn];
 bool vis[maxn];

 void chk(int u, int v, int w) {
     int cost1 = dist[s1][u] + w + dist[v][t1], cost2 = dist[s2][u] + w + dist[v][t2], cost = cost1 + cost2 - w;
     if (cost1 <= l1 && cost2 <= l2) min_update(ans, cost);
     swap(u, v);
     cost2 = dist[s2][u] + w + dist[v][t2];
     cost = cost1 + cost2 - w;
     if (cost1 <= l1 && cost2 <= l2) min_update(ans, cost);
 }

 void bfs(int s) {
     queue<pii> q;
     q.push(make_pair(s, ));
     mem0(vis);
     vis[s] = true;
     while (!q.empty()) {
         pii hnode = q.front(); q.pop();
         int u = hnode.first, w = hnode.second;
         dist[s][u] = w;
         int sz = G[u].size();
         rep_up0(i, sz) {
             int v = G[u][i];
             if (!vis[v]) {
                 vis[v] = true;
                 q.push(make_pair(v, w + ));
             }
         }
     }
 }

 int main() {
     //freopen("in.txt", "r", stdin);
     cin >> n >> m;
     memset(dist, 0x3f, sizeof(dist));
     G.clear();
     G.resize(n);
     rep_up0(i, m) {
         int u, v;
         sint2(u, v);
         G.add(u, v);
         G.add(v, u);
     }
     sint3(s1, t1, l1);
     sint3(s2, t2, l2);
     rep_up1(i, n) {
         bfs(i);
     }
     if (dist[s1][t1] > l1 || dist[s2][t2] > l2) {
         puts("-1");
         return ;
     }
     ans = dist[s1][t1] + dist[s2][t2];
     rep_up1(i, n) {
         rep_up1(j, n) {
             chk(i, j, dist[i][j]);
         }
     }
     cout << m - ans << endl;
     return ;
 }