題目鏈接

BZOJ4144

題解

這題好妙啊，，orz

假設我們在一個非加油站點，那麽我們一定是從加油站過來的，我們剩余的油至少要減去這段距離
如果我們在一個非加油站點，如果我們到達不了任意加油站點，我們一定廢了
那麽我們在一個非加油站點，就一定可以到達最近的加油站，而由於我們剩余的油是要減去到加油站距離的，所以我們剩余的油一定是\(b - d\)，\(d\)表示到達最近加油站的距離。假如我們沒有那麽多油，我們一定可以開過去再回來，就有了

因此，我們在任意一個點的油量確定，兩點之間可以直達，當且僅當
①兩點間有邊
②\(w + d[u] + d[v] \le b\)

因此可以以此為新邊權跑最小生成樹，同時離線回答詢問

#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;
}

BZOJ4144 [AMPPZ2014]Petrol 【最短路 + 最小生成樹】