HDU 3830 - Checkers (LCA模型)

Checkers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 1774    Accepted Submission(s): 515

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.

 

很详细的题解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define LL long long
struct node
{
    LL x,y,z;
    LL dis;
};

LL check(node a,node b)
{
    if(a.x==b.x&&a.y==b.y&&a.z==b.z) return 1;
    return 0;
}

node bong(node &b)
{
    node a=b;
    LL dis=0;
    while(a.z-a.y!=a.y-a.x){
        LL len1 = a.y-a.x;
        LL len2 = a.z-a.y;
        if(len1<len2){
            LL c=(len2-1)/len1;
            a.x+=c*len1;
            a.y+=c*len1;
            dis+=c;
        }else{
            LL c=(len1-1)/len2;
            a.z-=c*len2;
            a.y-=c*len2;
            dis+=c;
        }
    }
    b.dis=dis;
    return a;
}


void sort_(node &a)
{
    if(a.x>a.y) swap(a.x, a.y);
    if(a.y>a.z) swap(a.y, a.z);
    if(a.x>a.y) swap(a.y, a.x);
}


node update(node b,LL k)
{
    node a=b;
    while(k){
        LL len1 = a.y-a.x;
        LL len2 = a.z-a.y;
        if(len1<len2){
            LL c=(len2-1)/len1;
            c=min(c,k);
            a.x+=c*len1;
            a.y+=c*len1;
            k-=c;
            a.dis-=c;
        }else{
            LL c=(len1-1)/len2;
            c=min(c,k);
            a.z-=c*len2;
            a.y-=c*len2;
            k-=c;
            a.dis-=c;
        }
    }
    return a;
}

int main()
{
    node a;
    node b;
    while(~scanf("%lld %lld %lld",&a.x,&a.y,&a.z)){
        scanf("%lld %lld %lld",&b.x,&b.y,&b.z);
        sort_(a);sort_(b);
        node aa=bong(a);
        node bb=bong(b);
        if(!check(aa,bb)){
            printf("NO\n");
            continue;
        }
        if(a.dis<b.dis){
            swap(a, b);
        }
        LL ans=a.dis-b.dis;
        a=update(a, a.dis-b.dis);
        LL l=0,r=a.dis;
        while(l<r){
            LL mid = (l+r)>>1;
            aa=update(a, mid);
            bb=update(b, mid);
            if(check(aa, bb)){
                r=mid;
            }else{
                l=mid+1;
            }
        }
        printf("YES\n");
        printf("%lld\n",2*r+ans);
    }
    return 0;
}