「luogu4366」最短路

Posted 2020-11-25 zsbzsb

「luogu4366」最短路

传送门
直接连边显然不行，考虑优化。
根据异或的结合律和交换律等优秀性质，我们每次只让当前点向只有一位之别的另一个点连边，然后就直接跑最短路。
注意点数会很多，所以用配对堆优化 ( ext{Dijkstra}) 即可。
参考代码：

#include <cstring>
#include <cstdio>
#include <ext/pb_ds/priority_queue.hpp>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}
using namespace __gnu_pbds;
const int _ = 1e6 + 10, __ = 3e6 + 10;

int tot, head[_], nxt[__], ver[__], w[__];
inline void Add_edge(int u, int v, int d)
{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v, w[tot] = d; }

int n, m, c, s, t, dis[_], vis[_];
struct node { int val, u; } ;
inline bool operator < (const node& x, const node& y) { return x.val > y.val; }
priority_queue < node > Q;

inline void Dijkstra() {
    memset(dis, 0x3f, sizeof dis);
    Q.push((node) { 0, s }), dis[s] = 0;
    while (!Q.empty()) {
        int u = Q.top().u; Q.pop();
        if (vis[u]) continue ; vis[u] = 1;
        for (rg int i = head[u]; i; i = nxt[i]) {
            int v = ver[i];
            if (dis[v] > dis[u] + w[i])
                dis[v] = dis[u] + w[i], Q.push((node) { dis[v], v });
        }
    }
}

int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n), read(m), read(c);
    for (rg int u, v, d; m--; )
        read(u), read(v), read(d), Add_edge(u, v, d);
    for (rg int i = 1; i <= 100000; ++i)
        for (rg int j = 0; j <= 16; ++j) Add_edge(i, i ^ (1 << j), (1 << j) * c);
    read(s), read(t);
    Dijkstra();
    printf("%d
", dis[t]);
    return 0;
}