[IOI 2018] Werewolf

[题目链接]

         https://www.luogu.org/problemnew/show/P4899

[算法] 

        建出原图的最小/最大生成树的kruskal重构树然后二维数点

        时间复杂度 : O((N+Q)logN)

[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define N 200010
#define M 400010
#define MAXLOG 20
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void read(T &x)
{
   T f = 1; x = 0;
   char c = getchar();
   for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
   for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
   x *= f;
}

struct info
{
    int x , y;
} a[M];
struct edge
{
    int to , nxt;
} ea[N << 2] , eb[N << 2];

int n , m , q , tot , timera , timerb , cnta , cntb;
int la[N << 1] , ra[N << 1] , fathera[N << 1][MAXLOG] , fatherb[N << 1][MAXLOG] , 
    deptha[N << 1] , depthb[N << 1] , rt[N << 1] , heada[N << 1] , headb[N << 1] ,
    lb[N << 1] , rb[N << 1] , fa[N << 1] , valuea[N << 1] , valueb[N << 1] , w[N << 1];

struct Presitent_Segment_Tree 
{
    int sz;
    Presitent_Segment_Tree()
    {
        sz = 0;
    }
    struct node
    {
        int lc , rc;    
        int cnt;
    } a[N * 60];
    inline void modify(int &now , int old , int l , int r , int x , int value)
    {
        now = ++sz;
        a[now].lc = a[old].lc , a[now].rc = a[old].rc;
        a[now].cnt = a[old].cnt + value;
        if (l == r) return;
        int mid = (l + r) >> 1;
        if (mid >= x) modify(a[now].lc , a[old].lc , l , mid , x , value);
        else modify(a[now].rc , a[old].rc , mid + 1 , r , x , value);
    }
    inline int query(int rt1 , int rt2 , int l , int r , int ql , int qr)
    {
        if (l == ql && r == qr)
            return a[rt1].cnt - a[rt2].cnt;
        int mid = (l + r) >> 1;
        if (mid >= qr) return query(a[rt1].lc , a[rt2].lc , l , mid , ql , qr);
        else if (mid + 1 <= ql) return query(a[rt1].rc , a[rt2].rc , mid + 1 , r , ql , qr);
        else return query(a[rt1].lc , a[rt2].lc , l , mid , ql , mid) + query(a[rt1].rc , a[rt2].rc , mid + 1 , r , mid + 1 , qr);
    }
} PST;
inline void addedgea(int u , int v)
{
    ++tot;
    ea[tot] = (edge){v , heada[u]};
    heada[u] = tot;
}
inline void addedgeb(int u , int v)
{
    ++tot;
    eb[tot] = (edge){v , headb[u]};
    headb[u] = tot;
}
inline void dfsa(int u , int par)
{
    fathera[u][0] = par;
    deptha[u] = deptha[par] + 1;
    for (int i = 1; i < MAXLOG; i++)
        fathera[u][i] = fathera[fathera[u][i - 1]][i - 1];
    la[u] = ++timera;
    for (int i = heada[u]; i; i = ea[i].nxt)
    {
        int v = ea[i].to;
        if (v == par) continue;
        dfsa(v , u);
    }
    ra[u] = timera;
}
inline void dfsb(int u , int par)
{
    fatherb[u][0] = par;
    depthb[u] = depthb[par] + 1;
    for (int i = 1; i < MAXLOG; i++)
        fatherb[u][i] = fatherb[fatherb[u][i - 1]][i - 1];
    lb[u] = ++timerb;
    for (int i = headb[u]; i; i = eb[i].nxt)
    {
        int v = eb[i].to;
        if (v == par) continue;
        dfsb(v , u);
    }
    rb[u] = timerb;
}
inline bool cmpa(info a , info b)
{
    return min(a.x , a.y) > min(b.x , b.y);
}
inline bool cmpb(info a , info b)
{
    return max(a.x , a.y) < max(b.x , b.y);
}
inline int getroot(int x)
{
    if (fa[x] == x) return x;
    else return fa[x] = getroot(fa[x]);
}
inline void kruskalA()
{
    int tot = 0;
    cnta = n;
    timera = 0;
    sort(a + 1 , a + m + 1 , cmpa);
    for (int i = 1; i <= 2 * n; i++) fa[i] = i;
    for (int i = 1; i <= m; i++)
    {
        int fu = getroot(a[i].x) , fv = getroot(a[i].y);
        if (fu != fv)
        {
            ++cnta;
            addedgea(cnta , fu);
            addedgea(cnta , fv);
            valuea[cnta] = min(a[i].x , a[i].y);
            fa[fu] = fa[fv] = fa[cnta] = cnta;
            ++tot;
        }
        if (tot == n - 1) break; 
    }    
    dfsa(cnta , 0);
}
inline void kruskalB()
{
    int tot = 0;
    cntb = n;
    timerb = 0;
    sort(a + 1 , a + m + 1 , cmpb);
    for (int i = 1; i <= 2 * n; i++) fa[i] = i;
    for (int i = 1; i <= m; i++)
    {
        int fu = getroot(a[i].x) , fv = getroot(a[i].y);
        if (fu != fv)
        {
            ++cntb;
            addedgeb(cntb , fu);
            addedgeb(cntb , fv);
            valueb[cntb] = max(a[i].x , a[i].y);
            fa[fu] = fa[fv] = fa[cntb] = cntb;
            ++tot;
        }
        if (tot == n - 1) break;
    }
    dfsb(cntb , 0);
}
inline bool query(int l1 , int r1 , int l2 , int r2)
{
    return PST.query(rt[r1] , rt[l1 - 1] , 1 , 2 * n , l2 , r2);
}

int main()
{
    
    read(n); read(m); read(q);
    for (int i = 1; i <= m; i++)
    {
        read(a[i].x);
        read(a[i].y);
        ++a[i].x; ++a[i].y;
    }
    kruskalA();
    kruskalB();
    for (int i = 1; i <= n; i++) w[la[i]] = lb[i];
    for (int i = 1; i <= 2 * n; i++) PST.modify(rt[i] , rt[i - 1] , 1 , 2 * n , w[i] , 1);
    for (int i = 1; i <= q; i++)
    {
        int s , e , l , r;
        read(s); read(e); read(l); read(r);
        ++s; ++e; ++l; ++r;
        if (s < l || e > r) 
        {
            puts("0");
            continue;
        }
        for (int i = MAXLOG - 1; i >= 0; i--)
            if (fathera[s][i] && valuea[fathera[s][i]] >= l)
                s = fathera[s][i];
        for (int i = MAXLOG - 1; i >= 0; i--)
            if (fatherb[e][i] && valueb[fatherb[e][i]] <= r)
                e = fatherb[e][i];
        if (query(la[s] , ra[s] , lb[e] , rb[e])) puts("1");
        else puts("0");
    }
    
    return 0;
}