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 = 1505;
const int inf = 1e9;

int head[N], tot, vis[N];
int Head[N], Tot;
int dis[5][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 = 0, p = 1; char c = getchar();
    for (; c > '9' || c < '0'; c = getchar())
        if (c == '-') p = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        X = X * 10 + c - '0';
    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[1][x] + e[i].w + dis[2][e[i].to] != dis[1][t1])
        return 0;
    return dis[3][x] + e[i].w + dis[4][e[i].to] == dis[3][t2];
}

void spfa(int S, int *b) {
    for (int i = 1; i <= n; ++i)
        b[i] = inf;
    q.push(S);
    b[S] = 0;
    for (int u; !q.empty(); ) {
        u = q.front(); q.pop();
        vis[u] = 0;
        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] = 1, q.push(nt);
        }
    }
}

void bfs() {
    for (int u = 1; 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] != -1)
        return f[u];
    int tmp = 0;
    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 = 1; i <= m; ++i) {
        int u = rd, v = rd, w = rd;
        add(u, v, w); add(v, u, w);
    }
    spfa(s1, dis[1]); spfa(t1, dis[2]); spfa(s2, dis[3]); spfa(t2, dis[4]);
    bfs(); int ans = 0;
    memset(f, -1, sizeof(f));
    for (int i = 1; i <= n; ++i)
        ans = max(ans, dp(i));
    
    memset(f, -1, sizeof(f));
    memset(Head, 0, sizeof(Head));
    Tot = 0;
    swap(s2, t2);
    spfa(s2, dis[3]); spfa(t2, dis[4]);
    bfs();
    for (int i = 1; i <= n; ++i)
        ans = max(ans, dp(i));
    printf("%d
", ans);
}