bzoj2002: [Hnoi2010]Bounce 弹飞绵羊 lct

某天,Lostmonkey发明了一种超级弹力装置,为了在他的绵羊朋友面前显摆,他邀请小绵羊一起玩个游戏。游戏一开始,Lostmonkey在地上沿着一条直线摆上n个装置,每个装置设定初始弹力系数ki,当绵羊达到第i个装置时,它会往后弹ki步,达到第i+ki个装置,若不存在第i+ki个装置,则绵羊被弹飞。绵羊想知道当它从第i个装置起步时,被弹几次后会被弹飞。为了使得游戏更有趣,Lostmonkey可以修改某个弹力装置的弹力系数,任何时候弹力系数均为正整数。
题解:lct操作,对于可以从i一步到i+a[i]我们连一条边(如果i+a[i]>n那么就变成n+1),问题就变成了,i到n+1的最短路,每次更新先把原来的边cut,然后link上新边即可,(刚开始数组开小了一直t,还以为lct写搓了= = )

/**************************************************************
    Problem: 2002
    User: walfy
    Language: C++
    Result: Accepted
    Time:1740 ms
    Memory:6772 kb
****************************************************************/
 
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)
 
using namespace std;
 
const double eps=1e-6;
const int N=200000+10,maxn=5000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;
 
struct LCT{
    int fa[N],ch[N][2],rev[N],sz[N],q[N];
    inline bool isroot(int x){return ch[fa[x]][0]!=x&&ch[fa[x]][1]!=x;}
    inline void pushup(int x){sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+1;}
    inline void pushdown(int x)
    {
        if(rev[x])
        {
            rev[x]=0;swap(ch[x][0],ch[x][1]);
            rev[ch[x][0]]^=1,rev[ch[x][1]]^=1;
        }
    }
    inline void Rotate(int x)
    {
        int y=fa[x],z=fa[y],l,r;
        if(ch[y][0]==x)l=0,r=l^1;
        else l=1,r=l^1;
        if(!isroot(y))
        {
            if(ch[z][0]==y)ch[z][0]=x;
            else ch[z][1]=x;
        }
        fa[x]=z;fa[y]=x;fa[ch[x][r]]=y;
        ch[y][l]=ch[x][r];ch[x][r]=y;
        pushup(y);pushup(x);
    }
    inline void splay(int x)
    {
        int top=1;q[top]=x;
        for(int i=x;!isroot(i);i=fa[i])q[++top]=fa[i];
        for(int i=top;i;i--)pushdown(q[i]);
        while(!isroot(x))
        {
            int y=fa[x],z=fa[y];
            if(!isroot(y))
            {
                if((ch[y][0]==x)^(ch[z][0]==y))Rotate(x);
                else Rotate(y);
            }
            Rotate(x);
        }
    }
    inline void access(int x){for(int y=0;x;y=x,x=fa[x])splay(x),ch[x][1]=y,pushup(x);}
    inline void makeroot(int x){access(x),splay(x),rev[x]^=1;}
    inline int findroot(int x){access(x),splay(x);while(ch[x][0])x=ch[x][0];return x;}
    inline void split(int x,int y){makeroot(x),access(y),splay(y);}
    inline void cut(int x,int y){split(x,y);if(ch[y][0]==x)ch[y][0]=0,fa[x]=0;}
    inline void link(int x,int y){makeroot(x),fa[x]=y,splay(x);}
}lct;
int a[N];
inline const int read(){
    register int f=1,x=0;
    register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return f*x;
}
int main()
{
    int n=read();
    lct.sz[n+1]=1;
    for(register int i=1;i<=n;i++)
    {
        a[i]=read();
        a[i]=(i+a[i]<=n+1?i+a[i]:n+1);
        lct.fa[i]=a[i];
        lct.sz[i]=1;
    }
    int m=read();
    while(m--)
    {
        int op=read();
        if(op==1)
        {
            int x=read()+1;
            lct.makeroot(n+1);
            lct.access(x),lct.splay(x);
            printf("%d\n",lct.sz[x]-1);
        }
        else
        {
            int x=read()+1,y=read();
            lct.cut(x,a[x]);
            a[x]=(x+y<=n+1?x+y:n+1);
            lct.link(x,a[x]);
        }
    }
    return 0;
}
/********************
 
********************/
posted @ 2018-05-30 23:21  walfy  阅读(134)  评论(0编辑  收藏  举报