bzoj4144

最短路+最小生成树

有点忘了...

这题只要判断能不能就行了

具体做法是把所有加油站放到堆里然后跑dij,然后把边权w=d[u]+d[v]+w,跑最小生成树

对于点对(x,y)是否能到达只要判断最大瓶颈路的长度是否>b就行了

具体原因是因为对于一条边(u,v)，边权表示了经过这条边的最小油量。

然后最小生成树一边跑，一边离线就行了

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + ;
namespace IO
{
    const int Maxlen = N * ;
    char buf[Maxlen], *C = buf;
    int Len;
    inline void read_in()
    {
        Len = fread(C, , Maxlen, stdin);
        buf[Len] = '\0';
    }
    inline void fread(int &x)
    {
        x = ;
        int f = ;
        while (*C < '' || '' < *C) { if(*C == '-') f = -; ++C; }
        while ('' <= *C && *C <= '') x = (x << ) + (x << ) + *C - '', ++C;
        x *= f;
    }
    inline void fread(long long &x)
    {
        x = ;
        long long f = ;
        while (*C < '' || '' < *C) { if(*C == '-') f = -; ++C; }
        while ('' <= *C && *C <= '') x = (x << ) + (x << ) + *C - '', ++C;
        x *= f;
    }
    inline void read(int &x)
    {
        x = ;
        int f = ; char c = getchar();
        while(c < '' || c > '') { if(c == '-') f = -; c = getchar(); }
        while(c >= '' && c <= '') { x = (x << ) + (x << ) + c - ''; c = getchar(); }
        x *= f;
    }
    inline void read(long long &x)
    {
        x = ;
        long long f = ; char c = getchar();
        while(c < '' || c > '') { if(c == '-') f = -; c = getchar(); }
        while(c >= '' && c <= '') { x = (x << 1ll) + (x << 3ll) + c - ''; c = getchar(); }
        x *= f;
    }
} using namespace IO;
int rd()
{
    int x = , f = ; char c = getchar();
    while(c < '' || c > '') { if(c == '-') f = -; c = getchar(); }
    while(c >= '' && c <= '') { x = x *  + c - ''; c = getchar(); }
    return x * f;
}
int n, m, Q, cnt = , s;
int head[N], ans[N], c[N], fa[N], d[N], rank[N];
struct edge {
    int nxt, to, w;
} e[N << ];
struct Edge {
    int u, v, w;
    Edge() {}
    Edge(int u, int v, int w) : u(u), v(v), w(w) {}
    bool friend operator < (const Edge &a, const Edge &b) {
        return a.w < b.w;
    }
} E[N];
struct Query {
    int u, v, id, b;
    bool friend operator < (const Query &a, const Query &b) {
        return a.b < b.b;
    }
} que[N];
void link(int u, int v, int w)
{
    e[++cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].to = v;
    e[cnt].w = w;
}
int find(int x)
{
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}
int main()
{
    read_in();
    fread(n);
    fread(s);
    fread(m);
    memset(d, 0x7f7f, sizeof(d));
    priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;
    for(int i = ; i <= s; ++i)
    {
        fread(c[i]);
        d[c[i]] = ;
        q.push(make_pair(, c[i]));
    }
    for(int i = ; i <= m; ++i)
    {
        int u, v, w;
        fread(u);
        fread(v);
        fread(w);
        E[i] = Edge(u, v, w);
        link(u, v, w);
        link(v, u, w);
    }
    fread(Q);
    for(int i = ; i <= Q; ++i)
    {
        fread(que[i].u);
        fread(que[i].v);
        fread(que[i].b);
        que[i].id = i;
    }
    sort(que + , que + Q + );
    while(!q.empty())
    {
        pair<int, int> o = q.top();
        q.pop();
        int u = o.second;
        if(o.first > d[u]) continue;
        for(int i = head[u]; i; i = e[i].nxt) if(d[e[i].to] > d[u] + e[i].w)
        {
            d[e[i].to] = d[u] + e[i].w;
            q.push(make_pair(d[e[i].to], e[i].to));
        }
    }
    for(int i = ; i <= m; ++i) E[i].w += d[E[i].u] + d[E[i].v];
    sort(E + , E + m + );
    for(int i = ; i <= n; ++i) fa[i] = i;
    for(int i = , j = ; i <= Q; ++i)
    {
        while(j +  <= m && E[j + ].w <= que[i].b)
        {
            ++j;
            int u = find(E[j].u), v = find(E[j].v);
            if(u == v) continue;
            if(rank[u] < rank[v]) swap(u, v);
            rank[u] += rank[v];
            fa[u] = v;
        }
        ans[que[i].id] = find(que[i].u) == find(que[i].v);
    }
    for(int i = ; i <= Q; ++i) puts(ans[i] ? "TAK" : "NIE");
    return ;
}