HDU 3830 - Checkers(二分LCA)****

来源：互联网发布：迪杰斯特拉算法编辑：程序博客网时间：2024/05/01 18:22

题目：

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

题意：

给出直线上三个点的初始位置，和最终位置，每次可以选择一个点经过另一个点等间隔跳跃，但是不能经过两个点。求出是否能从初始状态到达最终状态。能的话输出最小步数。

思路：

将一个状态作为一个点，建二叉树。其根为状态能到达的三点等距的状态。

若题目给出的初始位置和最终位置都能到达同一个初始状态，则有答案。二分求解两个状态的LCA，答案即深度。

否则没有答案。

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

AC.

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;struct Node {    ll x, y, z;    ll dep;}s, t, rs, rt;void init(Node &a){    if(a.x > a.y) swap(a.x, a.y);    if(a.x > a.z) swap(a.x, a.z);    if(a.y > a.z) swap(a.y, a.z);}bool judge(Node a, Node b){    if(a.x == b.x && a.y == b.y && a.z == b.z)        return true;    return false;}ll Abs(ll x){    return x < 0? -x: x;}Node get_root(Node &a){    ll lxy, lyz;    Node r = a;    r.dep = 0;    ll dep = 0;    while(Abs(r.y - r.x) != Abs(r.z - r.y)) {        lxy = Abs(r.x - r.y); lyz = Abs(r.y - r.z);        if(lxy > lyz) {            ll c = (lxy-1)/lyz;            dep += c;            r.y -= c*lyz;            r.z -= c*lyz;        }        else {            ll c = (lyz - 1)/lxy;            dep += c;            r.x += c*lxy;            r.y += c*lxy;        }    }    a.dep = dep;    return r;}void update(Node &a, ll cnd){    ll cont = 0;    ll lxy, lyz;    while(cont < cnd) {        lxy = Abs(a.x - a.y);        lyz = Abs(a.y - a.z);        ll tmp = Abs(cnd-cont);        if(lxy > lyz) {            ll c = (lxy - 1)/lyz;            ll cc = min(c, tmp);            a.y -= cc*lyz;            a.z -= cc*lyz;            a.dep -= cc;            cont += cc;        }        else {            ll c = (lyz - 1)/lxy;            ll cc = min(c, tmp);            a.x += cc*lxy;            a.y += cc*lxy;            a.dep -= cc;            cont += cc;        }    }}ll solve(){    ll l = 0, r = s.dep;    ll ans;    Node ss, tt;    while(l <= r) {        ll mid = (l+r)/2;        ll dep = s.dep - mid;        ss = s; tt = t;        update(ss, dep);        update(tt, dep);        if(!judge(ss, tt)) {            r = mid - 1;        }        else l = mid+1;    }    return 2*(s.dep-r);}int main(){    //freopen("in", "r", stdin);    while(~scanf("%I64d%I64d%I64d", &s.x, &s.y, &s.z)) {        scanf("%I64d%I64d%I64d", &t.x, &t.y, &t.z);        init(s);        init(t);        rs = get_root(s), rt = get_root(t);        if(!judge(rs, rt)) {            printf("NO\n");            continue;        }        ll dep = abs(s.dep - t.dep);        if(s.dep > t.dep) {            update(s, dep);        }        else update(t, dep);        ll ans = solve();        printf("YES\n");        printf("%I64d\n", ans+dep);    }    return 0;}

0 0