题目链接

BZOJ4144

题解

这题好妙啊，，orz

假设我们在一个非加油站点，那么我们一定是从加油站过来的，我们剩余的油至少要减去这段距离

如果我们在一个非加油站点，如果我们到达不了任意加油站点，我们一定废了

那么我们在一个非加油站点，就一定可以到达最近的加油站，而由于我们剩余的油是要减去到加油站距离的，所以我们剩余的油一定是\(b - d\)，\(d\)表示到达最近加油站的距离。假如我们没有那么多油，我们一定可以开过去再回来，就有了

因此，我们在任意一个点的油量确定，两点之间可以直达，当且仅当

①两点间有边

②\(w + d[u] + d[v] \le b\)

因此可以以此为新边权跑最小生成树，同时离线回答询问

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<queue>

#include<cmath>

#include<map>

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define mp(a,b) make_pair<int,int>(a,b)

#define cls(s) memset(s,0,sizeof(s))

#define cp pair<int,int>

#define LL long long int

using namespace std;

const int maxn = 200005,maxm = 400005,INF = 2000000026;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

int n,s,m,Q,c[maxn];

int h[maxn],ne;

struct EDGE{int to,nxt,w;}ed[maxm];

inline void build(int u,int v,int w){

	ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;

	ed[++ne] = (EDGE){u,h[v],w}; h[v] = ne;

}

struct edge{int a,b,w;}e[maxn];

inline bool operator < (const edge& a,const edge& b){

	return a.w < b.w;

}

struct Que{int u,v,b,id;}q[maxn];

inline bool operator < (const Que& a,const Que& b){

	return a.b < b.b;

}

priority_queue<cp,vector<cp>,greater<cp> > qu;

int vis[maxn],d[maxn];

void dijkstra(){

	for (int i = 1; i <= n; i++) d[i] = INF;

	for (int i = 1; i <= s; i++)

		d[c[i]] = 0,qu.push(mp(0,c[i]));

	int u;

	while (!qu.empty()){

		u = qu.top().second; qu.pop();

		if (vis[u]) continue;

		vis[u] = true;

		Redge(u) if (!vis[to = ed[k].to] && d[u] + ed[k].w < d[to]){

			d[to] = d[u] + ed[k].w;

			qu.push(mp(d[to],to));

		}

	}

}

int ans[maxn],pre[maxn];

inline int find(int u){return u == pre[u] ? u : pre[u] = find(pre[u]);}

void kruskal(){

	REP(i,m) e[i].w = e[i].w + d[e[i].a] + d[e[i].b];

	sort(e + 1,e + 1 + m);

	int t = 1,fa,fb;

	REP(i,n) pre[i] = i;

	for (int i = 1; i <= m; i++){

		while (t <= Q && e[i].w > q[t].b){

			ans[q[t].id] = (find(q[t].u) == find(q[t].v));

			t++;

		}

		fa = find(e[i].a); fb = find(e[i].b);

		if (fa != fb) pre[fb] = fa;

	}

	while (t <= Q){

		ans[q[t].id] = (find(q[t].u) == find(q[t].v));

		t++;

	}

}

int main(){

	n = read(); s = read(); m = read();

	REP(i,s) c[i] = read();

	REP(i,m){

		e[i].a = read(); e[i].b = read(); e[i].w = read();

		build(e[i].a,e[i].b,e[i].w);

	}

	Q = read();

	REP(i,Q) q[i].u = read(),q[i].v = read(),q[i].b = read(),q[i].id = i;

	sort(q + 1,q + 1 + Q);

	dijkstra();

	kruskal();

	REP(i,Q) puts(ans[i] ? "TAK" : "NIE");

	return 0;

}