CF1117C

Description

题目大意是，船要从（x1, y1）行驶到（x2, y2），船可以上下左右行驶或呆在原位不动。同时还有风的影响，船的移动和风的影响可以叠加。问船行驶到终点的最少时间的多少？

思路

由于只要每次行驶和风相反的方向，就至少可以保证保持原位不动。所以船要么可以更靠近终点，要么保持不动，这就说明是单调的。所以二分所需的天数，判断只有风的影响下船的位置能否到达终点即可。
上限设置大一些，这样就能保证不能到达终点的情况的正确性。

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <cstring>
#include <string>
#include <deque>
#include <cmath>
#include <iomanip>
#include <cctype>
 
 
#define endl '
'
#define IOS std::ios::sync_with_stdio(0); 
#define FILE freopen("..//data_generator//in.txt","r",stdin),freopen("res.txt","w",stdout)
#define FI freopen("..//data_generator//in.txt","r",stdin)
#define FO freopen("res.txt","w",stdout)
#define md make_pair
#define pb push_back
#define mp make_pair
#define seteps(N) fixed << setprecision(N) 
 
typedef long long ll;
 
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
inline ll qmul(ll a, ll b, ll m) {
    ll res = 0;
    while(b) {
        if(b & 1) res = (res + a) % m;
        a = (a << 1) % m;
        b = b >> 1;
    }
    return res;
}
inline ll qpow(ll a, ll b, ll m) {
    ll res = 1;
    while(b) {
        if(b & 1) res = (res * a) % m;
        a = (a * a) % m;
        b = b >> 1;
    }
    return res;
}
inline ll inv(ll x, ll q) {
    return qpow(x, q - 2, q);
}
 
using namespace std;
/*-----------------------------------------------------------------*/
 
#define INF 0x3f3f3f3f
 
const int N = 1e6 + 10;
const double eps = 1e5;
char arr[N];
ll dx[N];
ll dy[N];
 
bool check(ll day, ll n, ll sx, ll sy, ll ex, ll ey) {
    ll x = ex - (sx + dx[day % n] + (day / n) * dx[n]);
    ll y = ey - (sy + dy[day % n] + (day / n) * dy[n]);
    return day >= abs(x) + abs(y);
}
 
int main() {
    IOS;
    //FI;
    int n;
    ll sx, sy;
    ll ex, ey;
    cin >> sx >> sy >> ex >> ey;
    cin >> n;
    for(int i = 1; i <= n; i++) {
        cin >> arr[i];
        if(arr[i] == 'U') dy[i]++;
        else if(arr[i] == 'D') dy[i]--;
        else if(arr[i] == 'L') dx[i]--;
        else if(arr[i] == 'R') dx[i]++;
        dx[i] += dx[i - 1];
        dy[i] += dy[i - 1];
    }
    ll l = 0, r = 1e18;
    while(l <= r) {
        ll mid = (l + r) / 2;
        if(check(mid, n, sx, sy, ex, ey)) r = mid - 1;
        else l = mid + 1;
    }
    if(l >= (ll)1e18) cout << -1 << endl;
    else cout << l << endl;
}