QOJ.ac
QOJ
| ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
|---|---|---|---|---|---|---|---|---|---|
| #506456 | #6422. Evil Coordinate | chanal | AC ✓ | 37ms | 3820kb | C++14 | 5.3kb | 2024-08-05 17:38:35 | 2024-08-05 17:38:36 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
int t;
void solve() {
long long mx,my;
cin>>mx>>my;
string str;
cin>>str;
map<char,int> mp;
for(int i=0; i<(int)str.size(); i++) {
mp[str[i]]++;//统计各个方向移动的指令数目
}
int end_x=mp['R']-mp['L'];//不要取绝对值就是算最后的坐标
int end_y=mp['U']-mp['D'];
if((mx==0&&my==0)||(end_x==mx&&end_y==my)) {
cout<<"Impossible"<<endl;
return;
}
/*0 -2
LDLRLLDRRL
endx=-2 endy-2
*/
if(my==0) { //地雷在x轴上
//最后落在x轴上,在这种情况下必然不会出现mx=0,那么就应该讨论mx和endx的关系以及end_y之间的关系
if(mp['U']==0&&mp['D']==0) { //如果都等于0这种情况说明只能左右移动那么只需要反方向移动即可
if(mx>0&&end_x>mx) {
cout<<"Impossible"<<endl;
return;
} else if(mx<0&&end_x<mx) {
cout<<"Impossible"<<endl;
return;
} else {
if(mx>0) {
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
for(int i=0; i<mp['R']; i++) {
cout<<'R';
}
cout<<endl;
return;
} else {
for(int i=0; i<mp['R']; i++) {
cout<<'R';
}
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
cout<<endl;
return;
}
}
} else if(end_y!=my) {
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
for(int i=0; i<mp['D']; i++) {
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;
} else if((mp['U']!=0)&&end_x!=mx) {
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
for(int i=0; i<mp['R']; i++) {
cout<<'R';
}
for(int i=0; i<mp['D']; i++) {
cout<<'D';
}
cout<<endl;
return;
} else if(mp['D']!=0&&end_x!=mx) {
for(int i=0; i<mp['D']; i++) {
cout<<'D';
}
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
for(int i=0; i<mp['R']; i++) {
cout<<'R';
}
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
cout<<endl;
return;
} else if(end_y==my) {
for(int i=0; i<mp['D']; i++) {
cout<<'D';
}
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
for(int i=0; i<mp['R']; i++) {
cout<<'R';
}
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
cout<<endl;
return;
}
} else if(mx==0) { //如果地雷在y轴上
if(mp['L']==0&&mp['R']==0) { //如果都等于0这种情况说明只能左右移动那么只需要反方向移动即可
if(my>0&&end_y>my) {
cout<<"Impossible"<<endl;
return;
} else if(my<0&&end_y<my) {
cout<<"Impossible"<<endl;
return;
} else {
if(my>0) {
for(int i=0; i<mp['D']; i++) {
cout<<'D';
}
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
cout<<endl;
return;
} else {
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_x!=mx) {
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
for(int i=0; i<mp['R']; i++) {
cout<<'R';
}
for(int i=0; i<mp['D']; i++) {
cout<<'D';
}
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
cout<<endl;
return;
} else if((mp['L']!=0)&&end_x!=mx) {
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
for(int i=0; i<mp['D']; i++) {
cout<<'D';
}
for(int i=0; i<mp['R']; i++) {
cout<<'R';
}
cout<<endl;
return;
} else if(mp['R']!=0&&end_x!=mx) {
for(int i=0; i<mp['R']; i++) {
cout<<'R';
}
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
for(int i=0; i<mp['D']; i++) {
cout<<'D';
}
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
cout<<endl;
return;
} else if(end_x==mx) {
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
for(int i=0; i<mp['D']; i++) {
cout<<'D';
}
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
for(int i=0; i<mp['R']; i++) {
cout<<'R';
}
cout<<endl;
return;
}
} else if(end_x!=mx&&end_y!=my) { //如果两个最终的结果都不跟坐标相同
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
for(int i=0; i<mp['R']; i++) {
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_x!=mx&&end_y==my) {//若x不相同,但是y的坐标相同的话__地雷的点不在坐标轴上,但是在
for(int i=0; i<mp['L']; i++) {
cout<<'L';
}
for(int i=0; i<mp['R']; i++) {
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,
for(int i=0; i<mp['U']; i++) {
cout<<'U';
}
for(int i=0; i<mp['D']; i++) {
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;
}
}
int main() {
// 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: 3820kb
input:
5 1 1 RURULLD 0 5 UUU 0 3 UUU 0 2 UUU 0 0 UUU
output:
LLRRUUD UUU Impossible Impossible Impossible
result:
ok 5 cases
Test #2:
score: 0
Accepted
time: 29ms
memory: 3720kb
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:
UURRDD UUR Impossible Impossible Impossible RRUUUUUUD LRRRRDD UD Impossible LLLLRRDD LLRRUDDD Impossible UUDDDDLLR LL Impossible UUUDLRR Impossible Impossible Impossible LLLLLRRRDD Impossible RL Impossible Impossible Impossible Impossible Impossible LLLRRRRRUU LLLUD Impossible UUULDDD UUDDRR Impossi...
result:
ok 11109 cases
Test #3:
score: 0
Accepted
time: 37ms
memory: 3640kb
input:
11107 1 0 LLRLRURLR 1 0 LLRR 0 1 R 1 0 LLLRLRRR 1 0 RUL 0 1 UD 1 0 RLRLU 0 1 DDDUUUDU 1 0 RURRLLRLL 1 0 LRLR 1 0 ULR 0 1 R 0 1 DDUUUDR 0 1 UUDDUDDU 0 1 DDUUDU 1 0 RRLRLLRLRL 1 0 RLRRLL 1 0 LUR 1 0 U 1 0 LRRRLLLR 0 1 DRUUDDUDU 0 1 DUUDDUR 1 0 LRLRLR 0 1 UUDDDUDU 0 1 R 0 1 UDUDDU 0 1 DUUDUD 1 0 RRLRRR...
output:
ULLLLRRRR LLRR R LLLLRRRR ULR DU ULLRR DDDDUUUU ULLLLRRRR LLRR ULR R RDDDUUU DDDDUUUU DDDUUU LLLLLRRRRR LLLRRR ULR U LLLLRRRR RDDDDUUUU RDDDUUU LLLRRR DDDDUUUU R DDDUUU DDDUUU LLLLLRRRRR DDDDUUUU DDUU ULLLLRRRR DDUU LLLRRR ULR ULR U ULR LLLRRR LLLLLRRRRR U DDDUUU R LLLRRR RDDDDUUUU RDDDDUUUU LLLRRR ...
result:
ok 11107 cases