Luogu 2149 [SDOI2009]Elaxia的路线

感觉这题可以模板化。

听说spfa死了，所以要练堆优化dijkstra。

首先对$x_{1},y_{1},x_{2},y_{2}$各跑一遍最短路，然后扫一遍所有边看看是不是同时在两个点的最短路里面，如果是的话就把这条边加到一张新图中去，因为最短路一定没有环，所以最后造出来的这张新图一定是一个$DAG$，dp一遍求最长链即为答案。

考虑一下怎么判断一条边是否在最短路里，设这条边连接的两个点是$x$，$y$，边权是$v$，如果它在最短路里面，那么有$dis(x_{1}, x) + v + dis(y_{1}, y) == dis(x_{1}, y_{1})$并且$dis(x_{2}, x) + v + dis(y_{2}, y) == dis(x_{2}, y_{2})$，注意第二个条件中$x$和$y$可以交换。加边的时候注意维持一下$DAG$的形态，可以把$x$和$y$到$x_{1}$的距离小的向距离大的连边。

时间复杂度$O(nlogn)$，堆优化dij是瓶颈。

感觉写得很长。

Code：

#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
using namespace std;
typedef pair <int, int> pin;

const int N = ;
const int M = 3e6 + ;
const int inf = 0x3f3f3f3f;

int n, m, inx[M], iny[M], inv[M], deg[N], f[N], ans = ;
int c1, c2, c3, c4, tot = , head[N], dis[N], d[][N];
bool vis[N];

struct Edge {
    int to, nxt, val;
} e[M << ];

inline void add(int from, int to, int val) {
    e[++tot].to = to;
    e[tot].val = val;
    e[tot].nxt = head[from];
    head[from] = tot;
}

inline void read(int &X) {
    X = ; char ch = ; int op = ;
    for(; ch > ''|| ch < ''; ch = getchar())
        if(ch == '-') op = -;
    for(; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}

inline void swap(int &x, int &y) {
    int t = x; x = y; y = t;
}

priority_queue <pin> Q;
void dij(int st) {
    memset(dis, 0x3f, sizeof(dis));
    memset(vis, , sizeof(vis));
    Q.push(pin(dis[st] = , st));
    for(; !Q.empty(); ) {
        int x = Q.top().second; Q.pop();
        if(vis[x]) continue;
        vis[x] = ;
        for(int i = head[x]; i; i = e[i].nxt) {
            int y = e[i].to;
            if(dis[y] > dis[x] + e[i].val) {
                dis[y] = dis[x] + e[i].val;
                Q.push(pin(-dis[y], y));
            }
        }
    }
} 

inline void chkMax(int &x, int y) {
    if(y > x) x = y;
}

int dfs(int x) {
    if(vis[x]) return f[x];
    vis[x] = ;
    int res = ;
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        chkMax(res, dfs(y) + e[i].val);
    }
    return f[x] = res;
}

int main() {
    read(n), read(m), read(c1), read(c2), read(c3), read(c4);
    for(int i = ; i <= m; i++) {
        read(inx[i]), read(iny[i]), read(inv[i]);
        add(inx[i], iny[i], inv[i]), add(iny[i], inx[i], inv[i]);
    }

    dij(c1); memcpy(d[], dis, sizeof(d[]));
    dij(c2); memcpy(d[], dis, sizeof(d[]));
    dij(c3); memcpy(d[], dis, sizeof(d[]));
    dij(c4); memcpy(d[], dis, sizeof(d[]));

/*    for(int i = 1; i <= n; i++)
        printf("%d ", d[0][i]);
    printf("\n");
    for(int i = 1; i <= n; i++)
        printf("%d ", d[1][i]);
    printf("\n");
    for(int i = 1; i <= n; i++)
        printf("%d ", d[2][i]);
    printf("\n");
    for(int i = 1; i <= n; i++)
        printf("%d ", d[3][i]);
    printf("\n");   */

    tot = ; memset(head, , sizeof(head));
    for(int i = ; i <= m; i++) {
        int x = inx[i], y = iny[i], v = inv[i];
        if(d[][x] + v + d[][y] == d[][c2])
            if(d[][y] + v + d[][x] == d[][c4] || d[][x] + v + d[][y] == d[][c4]) {
                if(d[][x] < d[][y]) {
                    add(x, y, v);
                    deg[y]++;
                } else {
                    add(y, x, v);
                    deg[x]++;
                }
            }

        swap(x, y);
        if(d[][x] + v + d[][y] == d[][c2])
            if(d[][y] + v + d[][x] == d[][c4] || d[][x] + v + d[][y] == d[][c4]) {
                if(d[][x] < d[][y]) {
                    add(x, y, v);
                    deg[y]++;
                } else {
                    add(y, x, v);
                    deg[x]++;
                }
            }
    }

    memset(vis, , sizeof(vis));
    for(int i = ; i <= n; i++)
        if(deg[i] ==  && !vis[i]) dfs(i);

/*    for(int i = 1; i <= n; i++)
        printf("%d ", f[i]);
    printf("\n");   */

    for(int i = ; i <= n; i++)
        chkMax(ans, f[i]);

    printf("%d\n", ans);
    return ;
}