hdu3830(lca + 二分)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3830

题意: 有三个点 a, b, c, 对于其中任意一点 x 可以跨过一个点移动到另一个位置, 当且仅当移动前后的 x 与其所跨越的点的距离相等 .给出两组点, 问其能否相互到达, 若能并输出最少需要移动多少步 .

思路: http://www.cnblogs.com/scau20110726/archive/2013/06/14/3135024.html

代码:

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #define ll long long
 using namespace std;

 struct node{
     ll x, y, z, d;
 };

 void f(node &s){
     ll cnt[];
     cnt[] = s.x;
     cnt[] = s.y;
     cnt[] = s.z;
     sort(cnt, cnt + );
     s.x = cnt[];
     s.y = cnt[];
     s.z = cnt[];
 }

 bool equal(node a, node b){//判断两个节点是否想等
     if(a.x == b.x && a.y == b.y && a.z == b.z) return true;
     return false;
 }

 node get_root(node &root){//这里要记录一下深度
     ll r;
     node cnt = root;
     ll p = cnt.y - cnt.x;
     ll q = cnt.z - cnt.y;
     while(p != q){
         if(q > p){ //右边大于左边
             r = (q - ) / p;
             cnt.x += r * p;
             cnt.y += r * p;
         }else{
             r = (p - ) / q;
             cnt.z -= r * q;
             cnt.y -= r * q;
         }
         root.d += r;
         f(cnt); //注意每个节点都要维护
         p = cnt.y - cnt.x;
         q = cnt.z - cnt.y;
     }
     return cnt;
 }

 node get_pre(node cnt, ll step){ //返回cnt向上移动step步后到达的位置
     ll p, q, r;
     while(step > ){
         p = cnt.y - cnt.x;
         q = cnt.z - cnt.y;
         if(q > p){
             r = (q - ) / p;
             if(r > step) r = step;//注意剩余步数不足的情况
             cnt.x += r * p;
             cnt.y += r * p;
         }else{
             r = (p - ) / q;
             if(r > step) r = step;
             cnt.z -= r * q;
             cnt.y -= r * q;
         }
         step -= r;
         f(cnt);
     }
     return cnt;
 }

 ll solve(node s, node e){
     ll l = , r = s.d < e.d ? s.d : e.d;
     ll cnt = r;
     while(l <= r){
         ll mid = (l + r) >> ;
         ll gg = cnt - mid;// mid 为 lca 位置, gg 为移动步数
         if(equal(get_pre(s, gg), get_pre(e, gg))) l = mid + ;
         else r = mid - ;
     }
     return ((cnt - (l - )) << );// l - 1 为lca位置
 }

 int main(void){
     node s, e;
     while(~scanf("%lld%lld%lld%lld%lld%lld", &s.x, &s.y, &s.z, &e.x, &e.y, &e.z)){
         s.d = e.d = ;
         f(s);
         f(e);
         if(!equal(get_root(s), get_root(e))){ //根节点不同无法到达
             puts("NO");
             continue;
         }
         ll d = abs(s.d - e.d);
         if(s.d > e.d) s = get_pre(s, d); //使两者处于同一深度
         else e = get_pre(e, d);
         int sol = solve(s, e);
         puts("YES");
         printf("%lld\n", d + sol);
     }
     return ;
 }