BZOJ 3489 A simple rmq problem

可持久化樹套樹

發現這是一個多維偏序，可持久化樹套樹即可。

#include<cstdio>
#include<algorithm> 
#define N 100005
using namespace std;
namespace runzhe2000
{
    int read()
    {
        int r = 0; char c = getchar();
        for(; c < '0' || c > '9'; c = getchar());
        for(; c >='0' && c <='9'; r = r*10+c-'0', c = getchar());
        return r;
    }
    int n, m, a[N], lastvis[N], pre[N], next[N], end[N], rk[N];
    struct seg0
    {
        seg0 *ch[2]; int v;
    }mem0[N*350], *tot0, *null0;
    int query0(seg0 *x, int l, int r, int ql, int qr)
    {
        if(ql <= l && r <= qr) return x->v; int mid = (l+r)>>1, ret = 0;
        if(ql <= mid) ret = max(ret, query0(x->ch[0], l,mid,ql,qr));
        if(mid <  qr) ret = max(ret, query0(x->ch[1], mid+1,r,ql,qr));
        return ret;
    }
    void modi0(seg0 *x, seg0 *y, int l, int r, int p, int val)
    {
        x->v = max(x->v, val);  if(l == r){ return;} int mid = (l+r)>>1;
        if(p <= mid) 
        {
            seg0 *tmp = ++tot0; *tmp = *y->ch[0];
            modi0(tmp,y->ch[0],l,mid,p,val);
            x->ch[0] = tmp;
        }
        else 
        {
            seg0 *tmp = ++tot0; *tmp = *y->ch[1];
            modi0(tmp,y->ch[1], mid+1,r,p,val);
            x->ch[1] = tmp;
        }
    }
    struct seg1
    {
        seg1 *ch[2]; seg0 *v;
    }mem1[N*40], *tot1, *root[N], *null1;
    int query1(seg1 *x, int l, int r, int ql, int qr, int qql, int qqr)
    {
        if(ql <= l && r <= qr) return query0(x->v,0,n+1,qql,qqr); int mid = (l+r)>>1, ret = 0;
        if(ql <= mid) ret = max(ret, query1(x->ch[0], l,mid,ql,qr,qql,qqr));
        if(mid <  qr) ret = max(ret, query1(x->ch[1], mid+1,r,ql,qr,qql,qqr));
        return ret;
    }
    void modi1(seg1 *x, seg1 *y, int l, int r, int p, int p2, int val)
    {
        seg0 *tmp = ++tot0; *tmp = *y->v;
        modi0(tmp, y->v, 0, n+1, p2, val);
        x->v = tmp;
        if(l == r) return; int mid = (l+r)>>1;
        if(p <= mid) 
        {
            seg1 *tmp = ++tot1; *tmp = *y->ch[0];
            modi1(tmp,y->ch[0],l,mid,p,p2,val);
            x->ch[0] = tmp;
        }
        else 
        {
            seg1 *tmp = ++tot1; *tmp = *y->ch[1];
            modi1(tmp,y->ch[1], mid+1,r,p,p2,val);
            x->ch[1] = tmp;
        }
    }
    void init()
    {
        null0 = tot0 = mem0; null0->ch[0] = null0->ch[1] = null0; null0->v = 0;
        null1 = tot1 = mem1; null1->ch[0] = null1->ch[1] = null1; null1->v = null0;
        root[0] = null1;
    }
    bool cmp(int a, int b){return pre[a] < pre[b];}
    void main()
    {
        n = read(), m = read();
        for(int i = 1; i <= n; i++) 
        {
            pre[i] = lastvis[a[i] = read()];
            lastvis[a[i]] = i;
        }
        for(int i = 1; i <= n; i++) lastvis[a[i]] = n+1;
        for(int i = n; i; i--)
        {
            next[i] = lastvis[a[i]];
            lastvis[a[i]] = i;
            rk[i] = i;
        }
        init(); sort(rk+1, rk+1+n, cmp);
        for(int i = 1; i <= n; i++)
        {
            root[i] = ++tot1, *root[i] = *root[i-1]; end[pre[rk[i]]] = i;
            modi1(root[i], root[i-1], 0, n+1, next[rk[i]], rk[i], a[rk[i]]);
        }
        for(int i = 1; i <= n; i++) end[i] = max(end[i], end[i-1]);
        int ans = 0, x, y, l, r;
        for(int i = 1; i <= m; i++)
        {
            x = read(), y = read();
            l = min((x+ans)%n+1, (y+ans)%n+1);
            r = max((x+ans)%n+1, (y+ans)%n+1);
            printf("%d\n",ans = query1(root[end[l-1]],0,n+1,r+1,n+1,l,r));
        }
    }
}
int main()
{
    runzhe2000::main();
}