hdu3830 (二分+LCA)

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Checkers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 775    Accepted Submission(s): 228

Problem Description

Little X, Little Y and Little Z are playing checkers when Little Y is annoyed. So he wants to make the chessboard much bigger. Although Little Z insists the original version, Little X stands by Little Y. After they enlarge the chessboard, the chessboard turns to an infinite line. 
The chessboard is like the Number Axes now, with each integer point able to hold a checker. At initial status there are three checkers on three different integer points , and through the game there always are three checkers. Every time, they can choose a checker A to jump across a pivot checker B to a new position(but the distance between old A and B equals to new A and B, and there should be no other checkers except B in the range [old A, new A]).
After playing for a while, they wonder whether an given status a,b,c can be transferred to x,y,z. obeying the rules. Since the checkers are considered the same, it is unnecessary for a must jump to x. 

 

Input

The first line is a,b,c.
The second line is x,y,z.
They are all integers in range (-10^9, 10^9) and the two status are valid.

 

Output

The first line is YES or NO, showing whether the transfer can be achieved.
If it is YES, then output the least steps in the second line.

 

Sample Input

1 2 3 0 3 5

 

Sample Output

YES 2

Hint

The middle checker jumps to position 0, and the status is 0 1 3 Then , the middle one jumps to 5.

 

Source

2011 Multi-University Training Contest 1 - Host by HNU

 

题意：

给出三个点数轴上的起始为止与终止位置，问能否通过不断的跳跃从起始位置到终止位置(ps:不需要一一对应)，每次跳跃时不得跨过两个点，若能，则给出最少需要几次操作

分析：

不妨设a<b<c,x<y<z。

由于一次不能跨越两个点。若b-a>c-b，则只能进行(1)b->2a-b-a或者(2)b->2c-b或者(3)a->2b-a;

若b-a<c-b，则只能进行(1)b->2a-b-a或者(2)b->2c-b或者(3)c->2b-c;

若b-a=c-b，则只能进行(1)b->2a-b-a或者(2)b->2c-b;

我们可以发现，若每次选取(1)(2)操作，则三个数在数轴上的跨度会不断变大，我们可以视为其分别向左儿子和右儿子移动，

若选取(3)操作，则三个数在数轴上的跨度会不断变小，对此，我们可以将其视为向父亲结点移动(因为对于每种状态，其对应的(3)操作是唯一的，这恰好对应这树的父亲结点是唯一的)。

通过一定次数的向父亲结点移动，我们可以发现最终一定会变成第三种状态。也就是对于判断yes还是no，我们只需要判断其所对应的根节点是否相同，若不同，则必定不能从初始状态转移到最终状态。

那么，如何快速地求出某一状态进行若干操作之后的状态呢？

我们假设a<b<c，并且c-b>a-b。那么，一次(3)操作之后会变成b,2b-a,c。也就是相当于给a和b同时加上了b-a。那么对于这样需要移动几次呢？(c-b)/(b-a)？

这是不对的，例如1，2，4。若移动二次，则会变成3,4,4。也就是我们要避免b-a=c-b的发生。即对于(c-b)%(b-a)==0时，我们需要少移一次。

于是，利用向下取整的特性，我们可以发现只要移(c-b-1)/(b-a)次。

如此，不断往复，我们就可以快速地求出移动若干次的状态，然后二分搞一下就行了

//#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>
using namespace std;
#define XINF INT_MAX
#define INF 0x3FFFFFFF
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI;
#define MAXN 10
ll a[MAXN],b[MAXN];
ll ra[MAXN],rb[MAXN];
void gcd(ll *x,ll &d){
    ll l1=x[1]-x[0];
    ll l2=x[2]-x[1];
    ll t;
    while(l1!=l2){
        if(l1<l2){
            t=(l2-1)/l1;
            ll tmp=t*l1;
            x[0]+=tmp;
            x[1]+=tmp;
        }else{
            t=(l1-1)/l2;
            ll tmp=t*l2; 
            x[1]-=tmp;
            x[2]-=tmp;
        }
        d+=t;
        l1=x[1]-x[0];
        l2=x[2]-x[1];
        //gcd(x,d);
    }
}
void gao(ll *x,ll d){
    ll l1=x[1]-x[0];
    ll l2=x[2]-x[1];
    ll t;
    while(d>0){
        if(l1<l2){
            t=(l2-1)/l1;
            if(t>d)t=d;
            ll tmp=t*l1;
            x[0]+=tmp;
            x[1]+=tmp;
        }else {
            t=(l1-1)/l2;
            if(t>d)t=d;
            ll tmp=t*l2;
            x[1]-=tmp;
            x[2]-=tmp;
        }
        d-=t;
        l1=x[1]-x[0];
        l2=x[2]-x[1];
    }
}
    
int main()
{
    ios::sync_with_stdio(false);
    while(scanf("%I64d%I64d%I64d",&a[0],&a[1],&a[2])!=EOF){
        for(int i=0;i<3;i++)scanf("%I64d",&b[i]);
        sort(a,a+3);
        sort(b,b+3);
        for(int i=0;i<3;i++)ra[i]=a[i];
        for(int i=0;i<3;i++)rb[i]=b[i];
        ll d1=0,d2=0;
        bool flag=0;
        for(int i=0;i<2;i++)if(ra[i]==ra[i+1])flag=1;
        for(int i=0;i<2;i++)if(rb[i]==rb[i+1])flag=1;
        if(flag){
            printf("NO
");
            continue;
        }
        gcd(ra,d1);
        gcd(rb,d2);
        flag=0;
        for(int i=0;i<3;i++)
            if(ra[i]!=rb[i])flag=1;
        if(flag){
            printf("NO
");
            continue;
        }
        printf("YES
");
        ll ans = 0;
        ll tmp = 0;
        if(d1>d2)gao(a,d1-d2);
        else gao(b,d2-d1);
        ll l=0,r=min(d1,d2);
        while(l<=r){
            ll mid=(l+r)>>1;
            for(int i=0;i<3;i++)ra[i]=a[i];
            for(int i=0;i<3;i++)rb[i]=b[i];
            gao(ra,mid);
            gao(rb,mid);
            bool flag=0;
            for(int i=0;i<3;i++)
                if(ra[i]!=rb[i])flag=1;
            if(!flag){
                tmp=mid;
                r=mid-1;
            }else {
                l=mid+1;
            }
        }
        printf("%I64d
",tmp*2+max(d1-d2,d2-d1));
    }    
    return 0;
}