SRM 507(2-1000pt)

DIV2 1000pt

题意：在一个长度无限的数轴上移动一个方块，每次可以向左或者向右移动距离x，只要x为完全平方数。数轴上有一些坑，如果方块移动到坑上则方块会掉进坑中，不能再被移动。给整数s,e，和所有坑的位置hol[i]，求最少多少步能够将方块从s点移动到e点，若不能从s移动到e，则返回-1。1 <= s, e, hol[i]  <= 100000，且他们互不相同。

　　　注意，若s = 1, e = 5, hol[0] = 3，则答案为-1。

解法：首先，从从s移动到e和从e移动到s的需要的步数是一样的，所以可以不妨设s < e。

　　　分为三种情况：存在一个i使得s < hol[i] < e，则答案一定为-1。

　　　若存在一个hol[i] < s和一个hol[j] > e，则将无限的数轴变为有限的，且1 <= hol[i] <= 100000，所以只需要一个建图用BFS做即可。

　　　

　　　然后先证明， 如果不存在hol[i] < s存在hol[j] > e，和不存在hol[i] < s且不存在hol[j] > e是等价的情况。这个很容易想明白，比如，从10转化到21，可以10 -> -15 -> 21，也可以10 ->46 -> 21。

　　　所以，剩下最后一种情况等价于，数轴上没有坑。设d = e - s，则无论如何，三步之内一定能从s走到e。

　　　若d为完全平方数，一步即可。

　　　若d = x*x + y*y，则两步即可。(枚举判断)

　　　若d为奇数，则d = 1 * d = (a+b) * (a-b)，即a = (d+1)/2，b = (d-1)/2，则a*a - b*b = d，两步即可。

　　　若d为偶数。则如果d为4的倍数，同样可以求出a和b使得d=a*a-b*b，两步即可。若d不为4的倍数，则d = 1 + (d-1)，三步即可。

　　　最后这种数轴上没有坑的情况，不能仅通过扩大有限数轴的范围然后用BFS的方法来做，因为范围会大到超时，而且也不容易判断具体范围要扩大到多大。

tag:math, BFS, good

// BEGIN CUT HERE
/*
 * Author:  plum rain
 * score :
 */
/*

 */
// END CUT HERE
#line 11 "CubeRoll.cpp"
#include <sstream>
#include <stdexcept>
#include <functional>
#include <iomanip>
#include <numeric>
#include <fstream>
#include <cctype>
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <queue>
#include <bitset>
#include <list>
#include <string>
#include <utility>
#include <map>
#include <ctime>
#include <stack>

using namespace std;

#define CLR(x) memset(x, 0, sizeof(x))
#define CLR1(x) memset(x, -1, sizeof(x))
#define PB push_back
#define SZ(v) ((int)(v).size())
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false)

typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long int64;
typedef pair<int, int> pii;

const double eps = 1e-8;
const double PI = atan(1.0)*4;
const int maxint = 2139062143;
const int maxx = 320;

bool v[100005];
pii an[100005];

bool ok(int a)
{
    int x = sqrt(a + 0.0);
    if ((x-1)*(x-1) == a) return 1;
    if (x * x == a) return 1;
    if ((x+1)*(x+1) == a) return 1;
    return 0;
}

int BFS (int l, int r, int s, int e)
{
    CLR (v);
    v[s] = 1;
    int il = 0, ir = 0;
    pii t; t.second = 1;
    for (int i = 1; i < maxx; ++ i){
        if (s + i*i < r){
            v[s+i*i] = 1;
            t.first = s + i*i; an[ir++] = t;
            continue;
        }
        else break;
    }
    for (int i = 1; i < maxx; ++ i){
        if (s - i*i > l){
            v[s - i*i] = 1;
            t.first = s - i*i; an[ir++] = t;
        }
        else break;
    }

    while (il < ir){
        pii tmp = an[il++];
        if (tmp.first == e) return tmp.second;

        pii t1 = tmp; ++ t1.second;
        for (int i = 1; i < maxx; ++ i){
            if (!v[tmp.first+i*i] && tmp.first + i*i < r){
                t1.first = tmp.first + i*i; an[ir++] = t1;
                v[t1.first] = 1;
            }
            if (tmp.first + i*i >= r) break;
        }
        for (int i = 1; i < maxx; ++ i){
            if (!v[tmp.first-i*i] && tmp.first - i*i > l){
                t1.first = tmp.first - i*i; an[ir++] = t1;
                v[t1.first] = 1;
            }
            if (tmp.first - i*i <= l) break;
        }
    }
    return -1;
}

int gao(int d)
{
    for (int i = 1; i < maxx; ++ i)
        for (int j = 1; j < maxx; ++ j)
            if (i*i + j*j == d) return 2;
    return 3;
}

class CubeRoll
{
    public:
        int getMinimumSteps(int s, int e, vector <int> hol){
            if (s > e) swap(s, e);
            bool tt = 0;
            int n = hol.size(), l = 0, r = 100005;
            for (int i = 0; i < n; ++ i){
                if (hol[i] < s) l = max(l, hol[i]);
                if (hol[i] > e) r = min(r, hol[i]);
                if (s < hol[i] && hol[i] < e) tt = 1;
            }
            
            if (tt) return -1;
            if (l && r != 100005) return BFS(l, r, s, e);

            int d = e - s;
            if (ok(d)) return 1;
            if (d & 1 || !(d % 4)) return 2;
            return gao(d);
        }
        
// BEGIN CUT HERE
    public:
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); }
    private:
    template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '"' << *iter << "","; os << " }"; return os.str(); }
    void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    void test_case_0() { int Arg0 = 5; int Arg1 = 1; int Arr2[] = {3}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = -1; verify_case(0, Arg3, getMinimumSteps(Arg0, Arg1, Arg2)); }
    void test_case_1() { int Arg0 = 36; int Arg1 = 72; int Arr2[] = {300, 100, 200, 400}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 1; verify_case(1, Arg3, getMinimumSteps(Arg0, Arg1, Arg2)); }
    void test_case_2() { int Arg0 = 10; int Arg1 = 21; int Arr2[] = {38,45}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 2; verify_case(2, Arg3, getMinimumSteps(Arg0, Arg1, Arg2)); }
    void test_case_3() { int Arg0 = 98765; int Arg1 = 4963; int Arr2[] = {10,20,40,30}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 2; verify_case(3, Arg3, getMinimumSteps(Arg0, Arg1, Arg2)); }
    void test_case_4() { int Arg0 = 68332; int Arg1 = 825; int Arr2[] = {99726,371,67,89210}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 2; verify_case(4, Arg3, getMinimumSteps(Arg0, Arg1, Arg2)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
//    freopen( "a.out" , "w" , stdout );    
    CubeRoll ___test;
    ___test.run_test(-1);
       return 0;
}
// END CUT HERE