AtCoder Beginner Contest 082
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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
andT
. - 1≤|s|≤8 000
- x and y are integers.
- |x|,|y|≤|s|
Input
Input is given from Standard Input in the following format:
sx y
Output
If the objective is achievable, print Yes
; if it is not, print No
.
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");}elseprintf("No\n"); }
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