AtCoder Beginner Contest 082 - D - FT Robot （set||DP）

最新推荐文章于 2021-07-30 19:59:19 发布

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本文介绍了一个关于二维平面上机器人路径规划的问题，通过解析指令序列来判断机器人是否能够达到指定坐标。采用类似母函数的DP实现方法，并使用set进行存储与更新。

D - FT Robot

Time limit : 2sec / Memory limit : 512MB

Score : 500 points

Problem Statement

A robot is put at the origin in a two-dimensional plane. Initially, the robot is facing in the positive x-axis direction.

This robot will be given an instruction sequence s. s consists of the following two kinds of letters, and will be executed in order from front to back.

F : Move in the current direction by distance 1.
T : Turn 90 degrees, either clockwise or counterclockwise.

The objective of the robot is to be at coordinates (x,y) after all the instructions are executed. Determine whether this objective is achievable.

Constraints

s consists of F and T.
1≤|s|≤8 000
x and y are integers.
|x|,|y|≤|s|

Input

Input is given from Standard Input in the following format:

s
x y

Output

If the objective is achievable, print Yes; if it is not, print No.

题意：

T转方向，可以任意逆或者顺，F往前走。

问你能不能到达x，y

POINT：

类似母函数的DP实现。

我用set实现了，比较慢

#include <set>
#include <map>
#include <math.h>
#include <vector>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
const double eps = 1e-8;
using namespace std;
 
set<int> x;
set<int> y;
set<int>::iterator it;
 
void doit(int a,int now)
{
	if(now==0) return;
	if(a==1){
		set<int> y_;
		y_.clear();
		for(it = y.begin();it!=y.end();it++){
			y_.insert(*it-now);
			y_.insert(*it+now);
		}
		y.clear();
		for(it = y_.begin();it!=y_.end();it++){
			y.insert(*it);
			y.insert(*it);
		}
	}else{
		set<int> x_;
		x_.clear();
		for(it = x.begin();it!=x.end();it++){
			x_.insert(*it-now);
			x_.insert(*it+now);
		}
		x.clear();
		for(it = x_.begin();it!=x_.end();it++){
			x.insert(*it);
			x.insert(*it);
		}
	}
}
 
int main()
{
	string s;
	cin>>s;
	int xx,yy;
	scanf("%d%d",&xx,&yy);
	int now=0;
	int i;
	for(i=0;i<s.length();i++){
		if(s[i]=='F'){
			now++;
		}else{
			break;
		}
	}
	x.insert(now+10000);
	y.insert(10000);
	int a=1;
	now=0;
	for(i++;i<s.length();i++){
		if(s[i]=='F'){
			now++;
		}else{
			doit(a,now);
			a^=1;
			now=0;
		}
	}
	doit(a,now);
	if(x.find(xx+10000)!=x.end()&&y.find(yy+10000)!=y.end()){
		printf("Yes\n");
	}else
		printf("No\n");
 
 
}