codeforces

链接:https://codeforces.com/problemset/problem/543/B
思路:首先求出任意两点间的最短路,因为边权都是1所以直接bfs就可以解决,然后我们考虑如果答案不是两个最短路值相加,中间必有一段是重合的,且起点终点分别是到这一段两个端点的最短路,所以直接枚举重合的部分即可。

代码:

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using namespace std;

const int maxn = 3010;
int d[maxn][maxn];
int n, m;
bool vis[maxn];
vector<int> G[maxn];

void (int s){
memset(vis, 0, sizeof(vis));
queue<int> q;
q.emplace(s);
vis[s] = 1;
while(!q.empty()){
int u = q.front();
q.pop();
for(auto &v : G[u]){
if(vis[v]) continue;
d[s][v] = d[s][u] + 1;
vis[v] = 1;
q.emplace(v);
}
}
}

int s1, t1, l1, s2, t2, l2;

int main(){
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cin >> n >> m;
for(int i = 1; i <= m; i++){
int u, v;
cin >> u >> v;
G[u].emplace_back(v);
G[v].emplace_back(u);
}
for(int i = 1; i <= n; i++){
bfs(i);
}
cin >> s1 >> t1 >> l1 >> s2 >> t2 >> l2;
if(d[s1][t1] > l1 || d[s2][t2] > l2){
cout << -1 << 'n';
return 0;
}
int res = d[s1][t1] + d[s2][t2];
for(int i = 1; i <= n; i++){
for(int j = i + 1; j <= n; j++){
if(d[s1][i] + d[i][j] + d[j][t1] <= l1 && d[s2][i] + d[i][j] + d[j][t2] <= l2)
res = min(res, d[s1][i] + d[j][t1] + d[s2][i] + d[j][t2] + d[i][j]);
if(d[s1][j] + d[i][j] + d[i][t1] <= l1 && d[s2][i] + d[i][j] + d[j][t2] <= l2)
res = min(res, d[s1][j] + d[i][t1] + d[s2][i] + d[j][t2] + d[i][j]);
if(d[s1][j] + d[i][j] + d[i][t1] <= l1 && d[s2][j] + d[i][j] + d[i][t2] <= l2)
res = min(res, d[s1][j] + d[i][t1] + d[s2][j] + d[i][t2] + d[i][j]);
if(d[s1][i] + d[i][j] + d[j][t1] <= l1 && d[s2][j] + d[i][j] + d[i][t2] <= l2)
res = min(res, d[s1][i] + d[j][t1] + d[s2][j] + d[i][t2] + d[i][j]);
}
}
cout << m - res << 'n';
return 0;
}