Sail（CodeForces - 298B ）

最新推荐文章于 2022-06-26 02:06:46 发布

starlet_kiss

最新推荐文章于 2022-06-26 02:06:46 发布

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CC 4.0 BY-SA版权

文章标签： codeforces 思维

本文链接：https://blog.youkuaiyun.com/starlet_kiss/article/details/84503896

本文探讨了极地熊在不同风向条件下，从起点到终点的最优航行策略。通过分析四个基本方向的移动，提出了一种算法来确定最早到达目的地的时间。该问题涉及到条件判断和路径选择，为读者提供了一个有趣且富有挑战性的编程实例。

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The polar bears are going fishing. They plan to sail from (sx, sy) to (ex, ey). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (x, y).

If the wind blows to the east, the boat will move to (x + 1, y).
If the wind blows to the south, the boat will move to (x, y - 1).
If the wind blows to the west, the boat will move to (x - 1, y).
If the wind blows to the north, the boat will move to (x, y + 1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (x, y). Given the wind direction for t seconds, what is the earliest time they sail to (ex, ey)?

Input
The first line contains five integers t, sx, sy, ex, ey (1 ≤ t ≤ 105, - 109 ≤ sx, sy, ex, ey ≤ 109). The starting location and the ending location will be different.

The second line contains t characters, the i-th character is the wind blowing direction at the i-th second. It will be one of the four possibilities: “E” (east), “S” (south), “W” (west) and “N” (north).

Output
If they can reach (ex, ey) within t seconds, print the earliest time they can achieve it. Otherwise, print “-1” (without quotes).

Examples
Input
5 0 0 1 1
SESNW
Output
4
Input
10 5 3 3 6
NENSWESNEE
Output
-1
Note
In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.

In the second sample, they cannot sail to the destination.
就是四个方向走，问问能不能走到终点，而且可以停在某个位置一秒。
一开始打算是模拟，但是一想，这玩意如何模拟啊，画了个图之后，才发现，如果目标点在起始点的西北部，那么给出的字符串里只要有足够的W和N就好了啊，分四种情况。
代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;

const int maxx=1e5+10;
char s[maxx];
int t,sx,sy,ex,ey;

int main()
{
	while(cin>>t>>sx>>sy>>ex>>ey)
	{
		cin>>s;
		int xx=ex-sx;
		int yy=ey-sy;
		if(xx==0&&yy==0)
		{
			cout<<0<<endl;
			continue;
		}
		int flag=0;
		if(xx<=0&&yy<=0)
		{
			for(int i=0;i<t;i++)
			{
				if(s[i]=='S'&&yy) yy++;
				if(s[i]=='W'&&xx) xx++;
				if(xx==0&&yy==0)
				{
					flag=1;
					cout<<i+1<<endl;
					break;
				}
			}
		}
		else if(xx<=0&&yy>=0)
		{
			for(int i=0;i<t;i++)
			{
				if(s[i]=='N'&&yy) yy--;
				if(s[i]=='W'&&xx) xx++;
				if(xx==0&&yy==0)
				{
					flag=1;
					cout<<i+1<<endl;
					break;
				}
			}
		}
		else if(xx>=0&&yy>=0)
		{
			for(int i=0;i<t;i++)
			{
				if(s[i]=='N'&&yy) yy--;
				if(s[i]=='E'&&xx) xx--;
				if(xx==0&&yy==0)
				{
					flag=1;
					cout<<i+1<<endl;
					break;
				}
			}
		}
		else if(xx>=0&&yy<=0)
		{
			for(int i=0;i<t;i++)
			{
				if(s[i]=='S'&&yy) yy++;
				if(s[i]=='E'&&xx) xx--;
				if(xx==0&&yy==0)
				{
					flag=1;
					cout<<i+1<<endl;
					break;
				}
			}
		}
		if(!flag) cout<<"-1"<<endl;
	}
}

努力加油a啊，(^o)/~