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。
1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 #include<cstdio> 5 #include<vector> 6 #include<stack> 7 #include<queue> 8 #include<cmath> 9 #include<map> 10 #include<set> 11 using namespace std; 12 typedef long long ll; 13 const int INF = 0x3f3f3f3f; 14 const int maxn=3500+5; 15 16 int n, m; 17 vector<int> G[maxn]; 18 int vis[maxn]; 19 int d[maxn][maxn]; 20 int s1,t1,l1,s2,t2,l2; 21 22 23 void bfs() 24 { 25 // memset(d,0,sizeof(d)); 26 for(int i=1;i<=n;i++) 27 { 28 memset(vis,0,sizeof(vis)); 29 queue<int> Q; 30 Q.push(i); 31 vis[i]=1; 32 while(!Q.empty()) 33 { 34 int u=Q.front(); Q.pop(); 35 for(int j=0;j<G[u].size();j++) 36 { 37 int v=G[u][j]; 38 if(!vis[v]) 39 { 40 vis[v]=1; 41 d[i][v]=d[i][u]+1; 42 Q.push(v); 43 } 44 } 45 } 46 } 47 } 48 49 50 bool judge(int s1, int t1, int s2, int t2, int i, int j) 51 { 52 return d[s1][i] + d[i][j] + d[j][t1] <= l1 && d[s2][i] + d[i][j] + d[j][t2] <= l2; 53 } 54 55 int main() 56 { 57 //freopen("in.txt","r",stdin); 58 scanf("%d%d",&n,&m); 59 for(int i=1;i<=m;i++) 60 { 61 int u,v; 62 scanf("%d%d",&u,&v); 63 G[u].push_back(v); 64 G[v].push_back(u); 65 } 66 67 scanf("%d%d%d",&s1,&t1,&l1); 68 scanf("%d%d%d",&s2,&t2,&l2); 69 70 bfs(); 71 72 int ans=d[s1][t1]+d[s2][t2]; 73 if(d[s1][t1]>l1 || d[s2][t2]>l2) puts("-1"); 74 else 75 { 76 for(int i=1;i<=n;i++) 77 for(int j=1;j<=n;j++) 78 { 79 if (judge(s1, t1, s2, t2, i, j)) 80 { 81 ans = min(ans, d[s1][i] + d[i][j] + d[j][t1] + d[s2][i] + d[j][t2]); 82 } 83 if (judge(t1, s1, t2, s2, i, j)) 84 { 85 ans = min(ans, d[t1][i] + d[i][j] + d[j][s1] + d[t2][i] + d[j][s2]); 86 } 87 if (judge(s1, t1, t2, s2, i, j)) 88 { 89 ans = min(ans, d[s1][i] + d[i][j] + d[j][t1] + d[t2][i] + d[j][s2]); 90 } 91 if (judge(t1, s1, s2, t2, i, j)) 92 { 93 ans = min(ans, d[t1][i] + d[i][j] + d[j][s1] + d[s2][i] + d[j][t2]); 94 } 95 } 96 printf("%d ",m-ans); 97 } 98 return 0; 99 }