Codeforces 1149D - Abandoning Roads

#include <bits/stdc++.h>
using namespace std;

inline int read() {
    int out = 0;
    register char cc = getchar();
    while (cc < ‘0‘ || cc > ‘9‘) cc = getchar();
    while (cc >= ‘0‘ && cc <= ‘9‘) out = (out << 3) + (out << 1) + (cc ^ 48), cc = getchar();
    return out;
}

int N, M, a, b, x, y, z, dist[71][1 << 18], check[71][1 << 18], ans[71];
int Id[80], tim;
queue<int> q1, q2, s1, s2;

int head[210], nxt[410], to[410], edge[410], cnt;

void add(int u, int v, int w) {
    to[++cnt] = v;
    edge[cnt] = w;
    nxt[cnt] = head[u];
    head[u] = cnt;
}

int fa[80], tot[80];

int find(int k) {
    return fa[k] == k ? k : fa[k] = find(fa[k]);
}

void Merge(int x, int y) {
    int rx = find(x), ry = find(y);
    if (rx != ry) fa[rx] = ry, tot[ry] += tot[rx];
}

int main() {
    N = read(), M = read(), a = read(), b = read();
    if (a > b) swap(a, b); // 保证 a < b 
    for (int i = 1; i <= N; i++) fa[i] = i, Id[i] = -1, tot[i] = 1, ans[i] = 0x3f3f3f3f;
    for (int i = 1; i <= M; i++) {
        x = read(), y = read(), z = read();
        if (z == a) Merge(x, y); // 用a边合并联通块 
        add(x, y, z);
        add(y, x, z);
    }
    for (int i = 1; i <= N; i++)
        if (Id[i] == -1 && tot[find(i)] > 3) {
            for (int j = 1; j <= N; j++) if (find(i) == find(j)) Id[j] = tim;
            ++tim;
        } // 给联通块标号(从0~tim-1) 
    for (register int i = 1; i <= N; i++)
        for (register int S = 0; S < (1 << tim); S++) dist[i][S] = 0x3f3f3f3f;
    dist[1][1 << Id[1]] = 0;
    q1.push(1); s1.push(Id[1] != -1 ? 1 << Id[1] : 0);
    while (!q1.empty() || !q2.empty()) {
        int u, S, S1, S2;
        x = y = 0x3f3f3f3f;
        if (!q1.empty()) x = q1.front(), S1 = s1.front();
        if (!q2.empty()) y = q2.front(), S2 = s2.front();
        if (x == 0x3f3f3f3f) u = y, S = S2, q2.pop(), s2.pop();
        else if (y == 0x3f3f3f3f) u = x, S = S1, q1.pop(), s1.pop();
        else if (dist[x][S1] < dist[y][S2]) u = x, S = S1, q1.pop(), s1.pop();
        else u = y, S = S2, q2.pop(), s2.pop();
        // 奇奇怪怪的最短路
        // 图上边权仅有 a, b
        // 将以a边更新得到的点与b边更新得到分装两个队列
        // 每次取较小的更新其它点
        // 据洛谷上巨佬说可以把复杂度降到O(N + M)
        check[u][S] = false;
        for (int i = head[u]; i; i = nxt[i]) {
            int v = to[i], w = edge[i];
            if (w == a && dist[v][S] > dist[u][S] + w) {
                dist[v][S] = dist[u][S] + w;
                if (!check[v][S]) q1.push(v), s1.push(S), check[v][S] = true;
            }
            if (w == b) {
                if (find(u) == find(v)) continue;
                if (Id[v] != -1 && ((1 << Id[v]) & S)) continue;
                int Next = (Id[v] != -1 ? S | (1 << Id[v]) : S);
                if (dist[v][Next] > dist[u][S] + w) {
                    dist[v][Next] = dist[u][S] + w;
                    if (!check[v][Next]) q2.push(v), s2.push(Next), check[v][Next] = true;
                }
            }  
        }
    }
    for (register int i = 1; i <= N; i++)
        for (register int S = 0; S < 1 << tim; S++)
            if (ans[i] > dist[i][S]) ans[i] = dist[i][S];
    for (int i = 1; i <= N; i++) cout << ans[i] << ‘ ‘;
    return 0;
}