luoguP4366 [Code+#4]最短路 最短路

好久没写过博客了....

本题还是挺有趣的(很水的最短路)

关键在于怎么优化这$n^2$条连边

通常，我们希望用一些边来替代一条边从而减小边集

那么，注意到异或操作可以拆分成按位运算，因此我们只需考虑$i$和每一位异或的结果连边即可

由于我们由$i$转移到$j$时，有可能中间节点$i \wedge t$是比$i$大的

因此，实际上我们应该带着$2^t$个点跑最短路，其中$2^t \geqslant n$

然后就没什么了...

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

extern inline char gc() {
    static char RR[], *S = RR + , *T = RR + ;
    if(S == T) fread(RR, , , stdin), S = RR;
    return *S ++;
}
inline int read() {
    int p = , w = ; char c = gc();
    while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
    while(c >= '' && c <= '') p = p *  + c - '', c = gc();
    return p * w;
}

#define sid 300050
#define eid 8005000
#define ll long long
#define ri register int

int n, m, c, cnp;
int vis[sid], cap[sid], nxt[eid], fee[eid], node[eid];

inline void addedge(int u, int v, int w) {
    nxt[++ cnp] = cap[u]; cap[u] = cnp;
    node[cnp] = v; fee[cnp] = w;
}

ll dis[sid];
struct P {
    int id; ll dis;
    friend bool operator < (P a, P b)
    { return a.dis > b.dis; }
};
priority_queue <P> q;

#define cur node[i]
void dij(int s, int t) {
    memset(dis, , sizeof(dis));
    dis[s] = ; q.push((P){ s,  });
    while(!q.empty()) {
        int id = q.top().id; ll di = q.top().dis; q.pop();
        if(vis[id]) continue; vis[id] = ;
        for(ri i = cap[id]; i; i = nxt[i])
        if(dis[cur] > di + fee[i]) dis[cur] = di + fee[i], q.push((P){ cur, dis[cur] });
    }
    printf("%lld\n", dis[t]);
}

int main() {
    n = read(); m = read(); c = read();
    for(ri i = ; i <= m; i ++) {
        int u = read(), v = read(), w = read();
        addedge(u, v, w);
    }
    int N = ;
    while(N <= n) N <<= ;
    for(ri i = ; i <= N; i ++)
    for(ri j = ; j <= N; j <<= ) addedge(i, i ^ j, j * c);
    int s = read(), t = read();
    dij(s, t);
    return ;
}