51nod 1444 破坏道路（最短路）

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1444

题意：

思路：

哇，思路爆炸。

因为每条边的权值都为1，所以可以直接用bfs来求出任意两个点之间的最短距离，复杂度为$O(n^2)$。

然后之后再暴力枚举一下，看看这两条路径之间是否会有重复的边。

 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cstdio>
 #include<vector>
 #include<stack>
 #include<queue>
 #include<cmath>
 #include<map>
 #include<set>
 using namespace std;
 typedef long long ll;
 const int INF = 0x3f3f3f3f;
 const int maxn=+;

 int n, m;
 vector<int> G[maxn];
 int vis[maxn];
 int d[maxn][maxn];
 int s1,t1,l1,s2,t2,l2;

 void bfs()
 {
    // memset(d,0,sizeof(d));
     for(int i=;i<=n;i++)
     {
         memset(vis,,sizeof(vis));
         queue<int> Q;
         Q.push(i);
         vis[i]=;
         while(!Q.empty())
         {
             int u=Q.front(); Q.pop();
             for(int j=;j<G[u].size();j++)
             {
                 int v=G[u][j];
                 if(!vis[v])
                 {
                     vis[v]=;
                     d[i][v]=d[i][u]+;
                     Q.push(v);
                 }
             }
         }
     }
 }

 bool judge(int s1, int t1, int s2, int t2, int i, int j)
 {
     return d[s1][i] + d[i][j] + d[j][t1] <= l1 && d[s2][i] + d[i][j] + d[j][t2] <= l2;
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     scanf("%d%d",&n,&m);
     for(int i=;i<=m;i++)
     {
         int u,v;
         scanf("%d%d",&u,&v);
         G[u].push_back(v);
         G[v].push_back(u);
     }

     scanf("%d%d%d",&s1,&t1,&l1);
     scanf("%d%d%d",&s2,&t2,&l2);

     bfs();

     int ans=d[s1][t1]+d[s2][t2];
     if(d[s1][t1]>l1 || d[s2][t2]>l2)  puts("-1");
     else
     {
         for(int i=;i<=n;i++)
         for(int j=;j<=n;j++)
         {
           if (judge(s1, t1, s2, t2, i, j))
             {
                 ans = min(ans, d[s1][i] + d[i][j] + d[j][t1] + d[s2][i] + d[j][t2]);
             }
             if (judge(t1, s1, t2, s2, i, j))
             {
                 ans = min(ans, d[t1][i] + d[i][j] + d[j][s1] + d[t2][i] + d[j][s2]);
             }
             if (judge(s1, t1, t2, s2, i, j))
             {
                 ans = min(ans, d[s1][i] + d[i][j] + d[j][t1] + d[t2][i] + d[j][s2]);
             }
             if (judge(t1, s1, s2, t2, i, j))
             {
                 ans = min(ans, d[t1][i] + d[i][j] + d[j][s1] + d[s2][i] + d[j][t2]);
             }
         }
         printf("%d\n",m-ans);
     }
     return ;
 }