Luogu2149 [SDOI2009]Elaxia的路线-最短路+拓扑排序

Solution

另外$ m <=5e5$。

两条最短路的 最长公共路径 一定是若干条连续的边， 并且满足拓扑序。

于是我们分别 正向 和反向走第二条路径，若该条边同时是两条最短路径上的边， 则加入边集。

最后拓扑 求最长链即可

Code

 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #include<queue>
 #define rd read()
 using namespace std;

 const int N = ;
 const int inf = 1e9;

 int head[N], tot, vis[N];
 int Head[N], Tot;
 int dis[][N], f[N];
 int n, m, s1, s2, t1, t2;

 queue<int> q;

 struct edge {
     int nxt, to, w;
 }e[N * N], E[N * N];

 int read() {
     int X = , p = ; char c = getchar();
     for (; c > '' || c < ''; c = getchar())
         if (c == '-') p = -;
     for (; c >= '' && c <= ''; c = getchar())
         X = X *  + c - '';
     return X * p;

 }

 void add(int u, int v, int w) {
     e[++tot].to = v;
     e[tot].nxt = head[u];
     e[tot].w = w;
     head[u] = tot;
 }

 void Add(int u, int v, int w) {
     E[++Tot].to = v;
     E[Tot].nxt = Head[u];
     E[Tot].w = w;
     Head[u] = Tot;
 }

 int jud(int x, int i) {
     if (dis[][x] + e[i].w + dis[][e[i].to] != dis[][t1])
         return ;
     return dis[][x] + e[i].w + dis[][e[i].to] == dis[][t2];
 }

 void spfa(int S, int *b) {
     for (int i = ; i <= n; ++i)
         b[i] = inf;
     q.push(S);
     b[S] = ;
     for (int u; !q.empty(); ) {
         u = q.front(); q.pop();
         vis[u] = ;
         for (int i = head[u]; i; i = e[i].nxt) {
             int nt = e[i].to;
             if (b[nt] <= b[u] + e[i].w)
                 continue;
             b[nt] = b[u] + e[i].w;
             if (!vis[nt])
                 vis[nt] = , q.push(nt);
         }
     }
 }

 void bfs() {
     for (int u = ; u <= n; ++u)
         for (int i = head[u]; i; i = e[i].nxt)
             if (jud(u, i)) Add(u, e[i].to, e[i].w);
 }

 int dp(int u) {
     if (f[u] != -)
         return f[u];
     int tmp = ;
     for (int i = Head[u]; i; i = E[i].nxt) {
         int nt = E[i].to;
         tmp = max(dp(nt) + E[i].w, tmp);
     }
     return f[u] = tmp;
 }

 int main()
 {
     n = rd; m = rd;
     s1 = rd; t1 = rd; s2 = rd; t2 = rd;
     for (int i = ; i <= m; ++i) {
         int u = rd, v = rd, w = rd;
         add(u, v, w); add(v, u, w);
     }
     spfa(s1, dis[]); spfa(t1, dis[]); spfa(s2, dis[]); spfa(t2, dis[]);
     bfs(); int ans = ;
     memset(f, -, sizeof(f));
     for (int i = ; i <= n; ++i)
         ans = max(ans, dp(i));

     memset(f, -, sizeof(f));
     memset(Head, , sizeof(Head));
     Tot = ;
     swap(s2, t2);
     spfa(s2, dis[]); spfa(t2, dis[]);
     bfs();
     for (int i = ; i <= n; ++i)
         ans = max(ans, dp(i));
     printf("%d\n", ans);
 }