「AMPPZ2014」Petrol

传送门：
这是一道bzoj权限题
Luogu团队题链接

解题思路

首先对于每一个点 \(x\) 预处理出 \(nr[x]\) 和 \(dis[x]\)，分别表示离 \(x\) 最近的加油站以及该段距离。
这个过程可以用多源 \(\text{Dijkstra}\) 处理。
然后对于每一条原图中的边 \((u, v, w)\)（\(nr[u]\ne nr[v]\)）
改为这样一条边：\((nr[u], nr[v], dis[u] + dis[v] + w)\)
然后只要离线用并查集维护一下连通性即可。

细节注意事项

• 最短路不要写挂啊

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#define rg register
#define pii pair < int, int >
using namespace std;
template < typename T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while (!isdigit(c)) f |= (c == '-'), c = getchar();
    while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
    s = f ? -s : s;
}

const int _ = 200010;
const int __ = 400010;

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 q, n, m, k, c[_], vis[_], dis[_], nr[_];
struct node{ int u, v, d, id; }ask[_], g[__], e[__]; int ans[_];

inline bool cmp(const node& x, const node& y) { return x.d < y.d; }

inline void Dijkstra() {
    static priority_queue < pii > Q;
    memset(dis, 0x3f, sizeof dis);
    for (rg int i = 1; i <= k; ++i)
        Q.push(make_pair(dis[c[i]] = 0, c[i])), nr[c[i]] = c[i];
    while (!Q.empty()) {
        int u = Q.top().second; 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], nr[v] = nr[u], Q.push(make_pair(-dis[v], v));
        }
    }
}

int fa[_];

inline int findd(int x) { return fa[x] == x ? x : fa[x] = findd(fa[x]); }

inline void merge(int x, int y) { fa[findd(x)] = findd(y); }

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.in", "r", stdin);
#endif
    read(n), read(k), read(m);
    for (rg int i = 1; i <= k; ++i) read(c[i]);
    for (rg int u, v, d, i = 1; i <= m; ++i)
        read(u), read(v), read(d), Add_edge(u, v, d), Add_edge(v, u, d), g[i] = (node) { u, v, d };
    Dijkstra();
    int cnt = 0;
    for (rg int u, v, i = 1; i <= m; ++i) {
        u = g[i].u, v = g[i].v;
        if (nr[u] != nr[v]) e[++cnt] = (node) { nr[u], nr[v], dis[u] + dis[v] + g[i].d };
    }
    read(q);
    for (rg int i = 1; i <= q; ++i)
        read(ask[i].u), read(ask[i].v), read(ask[i].d), ask[i].id = i;
    sort(ask + 1, ask + q + 1, cmp);
    sort(e + 1, e + cnt + 1, cmp);
    for (rg int i = 1; i <= n; ++i) fa[i] = i;
    for (rg int i = 1, j = 1; i <= q; ++i) {
        while (j <= cnt && e[j].d <= ask[i].d) merge(e[j].u, e[j].v), ++j;
        ans[ask[i].id] = findd(ask[i].u) == findd(ask[i].v);
    }
    for (rg int i = 1; i <= q; ++i) puts(ans[i] ? "TAK" : "NIE");
    return 0;
}

完结撒花 \(qwq\)

原文地址：https://www.cnblogs.com/zsbzsb/p/11745703.html