Educational Codeforces Round 60 (Rated for Div. 2)

Problem   Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship

Time Limit: 2000 mSec

Problem Description

Input

Output

The only line should contain the minimal number of days required for the ship to reach the point (x2,y2)(x2,y2).

If it's impossible then print "-1".

Sample Input

0 0
4 6
3
UUU

Sample Output

5

题解：第一感觉是bfs，然后大概背包一下之类的，但是数据范围不允许呀，做出这个题的关键点就是注意到运动独立性（呵呵），和如果在第x天能到，那么对于y >= x的都能到，因此就可以二分，固定了天数之后，风吹的距离就是固定的，并且可以用前缀和O(1)求，然后就是判断风吹到的点和目标点的哈密顿距离是否<=天数即可。

#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i = 1; i <= (n); i++)
#define sqr(x) ((x) * (x))

const int maxn = 100000 + 100;
const int maxm = 200000 + 100;
const int maxs = 10000 + 10;

typedef long long LL;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;

const LL unit = 1LL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-14;
const double inf = 1e15;
const double pi = acos(-1.0);
const int SIZE = 100 + 5;
const LL MOD = 1000000007;

LL sx, sy, ex, ey;
LL x[maxn], y[maxn];
LL n;
string str;

bool Judge(LL lim)
{
    LL m = lim / n, remain = lim % n;
    LL xx = sx + m * x[n] + x[remain];
    LL yy = sy + m * y[n] + y[remain];
    if (abs(xx - ex) + abs(yy - ey) <= lim)
        return true;
    return false;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    cin >> sx >> sy >> ex >> ey;
    cin >> n;
    cin >> str;
    LL len = str.size();
    for (LL i = 0; i < len; i++)
    {
        if (str[i] == 'U')
        {
            x[i + 1] = x[i];
            y[i + 1] = y[i] + 1;
        }
        else if (str[i] == 'D')
        {
            x[i + 1] = x[i];
            y[i + 1] = y[i] - 1;
        }
        else if (str[i] == 'L')
        {
            x[i + 1] = x[i] - 1;
            y[i + 1] = y[i];
        }
        else
        {
            x[i + 1] = x[i] + 1;
            y[i + 1] = y[i];
        }
    }
    LL le = 0, ri = LLONG_MAX / 10;
    LL ans = -1;
    while (le <= ri)
    {
        LL mid = (le + ri) / 2;
        if (Judge(mid))
        {
            ri = mid - 1;
            ans = mid;
        }
        else
        {
            le = mid + 1;
        }
    }
    cout << ans << endl;

    return 0;
}