bzoj2144

转载于 2017-11-02 07:46:00 发布 · 65 阅读

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原文链接：http://www.cnblogs.com/19992147orz/p/7770499.html

二分+lca

我们把向中间缩看成向上爬，向两边走看成向下爬，那么就相当于找出两个状态的lca，如果相邻的差是(a,b),a<b，那么向中间走就是(a,b-a)或(b-a,a),这个东西很像更相减损术，那么我们直接用(b-1)/a算出来要走的步数，然后继续递归求lca，直到走不了为止。先爬inf步判断是否有共同的祖先，然后将比较深的爬到同一高度，然后二分爬的步数，每次求lca就行了。

思路很奇妙啊

#include<bits/stdc++.h>
using namespace std;
struct data {
    int a[3];
    data() { memset(a, 0, sizeof(a)); }
    bool friend operator != (const data &a, const data &b) {
        for(int i = 0; i < 3; ++i) if(a.a[i] != b.a[i]) return true;
        return false;
    }
};
int dd, s1, s2;
int a[3], b[3];
data lca(int *a, int d) 
{
    data ret;
    int t1 = a[1] - a[0], t2 = a[2] - a[1];
    for(int i = 0; i < 3; ++i) ret.a[i] = a[i];
    if(t1 == t2) return ret;
    if(t1 < t2) 
    {
        int tmp = min(d, (t2 - 1) / t1);
        d -= tmp;
        dd += tmp;
        ret.a[0] += tmp * t1;
        ret.a[1] += tmp * t1;
    }
    else
    {
        int tmp = min(d, (t1 - 1) / t2);
        d -= tmp;
        dd += tmp;
        ret.a[2] -= tmp * t2;
        ret.a[1] -= tmp * t2;
    }
    return d ? lca(ret.a, d) : ret; 
}
int main()
{
    for(int i = 0; i < 3; ++i) scanf("%d", &a[i]);
    for(int i = 0; i < 3; ++i) scanf("%d", &b[i]);
    sort(a, a + 3);
    sort(b, b + 3);
    data t1 = lca(a, 1e9);
    s1 = dd;
    dd = 0;
    data t2 = lca(b, 1e9);
    s2 = dd;
    dd = 0;
    if(t1 != t2) 
    {
        puts("NO");
        return 0;
    }
    if(s1 < s2) 
    {
        swap(s1, s2);
        for(int i = 0; i < 3; ++i) swap(a[i], b[i]);
    }
    t1 = lca(a, s1 - s2);
    for(int i = 0; i < 3; ++i) a[i] = t1.a[i];
    int l = 0, r = 1e9, ans = 0; 
    while(r - l > 1)
    {
        int mid = (l + r) >> 1;
        if(lca(a, mid) != lca(b, mid)) l = mid;
        else r = ans = mid;
    } 
    if(ans && !(lca(a, ans - 1) != lca(b, ans - 1))) --ans;
    printf("YES\n%d\n", s1 - s2 + 2 * ans);
    return 0;
}