QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#506120#6422. Evil CoordinateAbclTL 0ms3548kbC++142.7kb2024-08-05 15:23:382024-08-05 15:23:38

Judging History

你现在查看的是最新测评结果

  • [2024-08-05 15:23:38]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:3548kb
  • [2024-08-05 15:23:38]
  • 提交

answer

#include <bits/stdc++.h>
#define int long long
using namespace std;
vector<char> ans;
int end_x,end_y;
unordered_map<char,int> mp;
int f=0;
int mx,my;
void dfs(int nowx,int nowy){
	if(nowx==end_x&&nowy==end_y){
		f=1;
		return;
	}
			if(nowx+1!=mx&&nowy!=my&&mp['R']){
				mp['R']--;
				ans.emplace_back('R');
				dfs(nowx+1,nowy);
				mp['R']++;ans.pop_back();
				if(f)return;
			}
			if(nowx-1!=mx&&nowy!=my&&mp['L']){
				mp['L']--;
				ans.emplace_back('L');
				dfs(nowx-1,nowy);
				mp['L']++;ans.pop_back();
				if(f)return;
			}
			if(nowx!=mx&&nowy+1!=my&&mp['U']){
				mp['U']--;
				ans.emplace_back('U');
				dfs(nowx,nowy+1);
				mp['U']++;ans.pop_back();
				if(f)return;	
			}
			if(nowx!=mx&&nowy-1!=my&&mp['D']){
				mp['D']--;
				ans.emplace_back('D');
				dfs(nowx,nowy-1);
				mp['D']++;ans.pop_back();
				if(f)return;	
			}

}
void solve() {
	cin>>mx>>my;
	string str;
	cin>>str;
	f=0;ans.clear();
	if(mx==0&&my==0) {
		cout<<"Impossible"<<endl;
		return;
	}
	mp.clear();
	for(int i=0; i<str.length(); i++) {
		mp[str[i]]++;//统计各个方向移动的指令数目
	}
	end_x=mp['R']-mp['L'];//不要取绝对值就是算最后的坐标
	end_y=mp['U']-mp['D'];
	if(mx==end_x&&my==end_y){
		cout<<"Impossible"<<endl;
		return;
	}
	if(end_x!=mx&&end_y!=my) { //如果两个最终的结果都不跟坐标相同
		dfs(0,0);
		for(auto it:ans)cout<<it;
		cout<<endl;
		return;
	} 
	
	else if(end_x!=mx&&end_y==my) {//若x不相同,但是y的坐标相同的话
	
			if(mx>0&&end_x>mx){
				cout<<"Impossible"<<endl;
				return;
			}
			else if(mx<0&&end_x<mx){
					cout<<"Impossible"<<endl;
					return;
			}
						
			for(int i=1;i<=min(mp['L'],mp['R']); i++) {
				cout<<'L'<<'R';
			}
			for(int i=min(mp['L'],mp['R'])+1; i<=max(mp['L'],mp['R']); i++) {
				if(mp['L']>mp['R'])
					cout<<'L';
				else cout<<'R';
			}
			for(int i=0; i<mp['U']; i++) {
				cout<<'U';
			}
			for(int i=0; i<mp['D']; i++) {
				cout<<'D';
			}
			cout<<endl;
			return;
		
	} 
	else if(end_y!=my&&end_x==mx) { //如果y最后不相同但是x相同,
			if(my>0&&end_y>my){
				cout<<"Impossible"<<endl;
				return;
			}
			else if(my<0&&end_y<my){
				cout<<"Impossible"<<endl;
				return;
			}
			
			for(int i=1; i<=min(mp['U'],mp['D']); i++) {
				cout<<'U'<<'D';
			}
			for(int i=min(mp['U'],mp['D'])+1;i<=max(mp['U'],mp['D']); i++){
				if(mp['U']>mp['D'])
					cout<<'U';
				else cout<<'D';
			}
			for(int i=0; i<mp['L']; i++) {
				cout<<'L';
			}
			for(int i=0; i<mp['R']; i++) {
				cout<<'R';
			}
			cout<<endl;
			return;
	} 
	
}
signed main() {
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int t;
	cin>>t;
	while(t--) {
		solve();
	}
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3548kb

input:

5
1 1
RURULLD
0 5
UUU
0 3
UUU
0 2
UUU
0 0
UUU

output:

LRLRUUD
UUU
Impossible
Impossible
Impossible

result:

ok 5 cases

Test #2:

score: -100
Time Limit Exceeded

input:

11109
6 0
RUDUDR
2 0
URU
0 0
UDRU
0 0
R
-1 1
LDUUDDRUUL
-1 5
RRUUUDUUU
-8 4
RRDRLDR
2 0
UD
0 0
UUDD
3 -2
LDDLLLRR
3 -2
LDRURLDD
1 0
RRL
-1 0
DUDDLLRDU
-4 0
LL
-1 -1
DLRLDLUDUR
1 4
URDULUR
0 0
DDUUDUDDDD
0 2
UU
1 0
RRULD
0 -2
LDLRLLDRRL
0 1
RLRLLRLUR
-3 0
RL
0 0
D
0 0
L
0 0
DDLRRUDRUD
0 0
DULU
2 0
RR...

output:

RRUUDD

Impossible
Impossible
Impossible
RRUUUUUUD

UD
Impossible
LRLRLLDD
LRLRUDDD
Impossible
UDUDDDLLR
LL
Impossible
UDUULRR
Impossible
Impossible
Impossible
LRLRLRLLDD
Impossible
LR
Impossible
Impossible
Impossible
Impossible
Impossible
LRLRLRRRUU

Impossible
LUUUDDD
UDUDRR
Impossible
LRRUUDD


L...

result: