第 45 屆國際大學生程式設計競賽（ICPC）亞洲區域賽（南京）E Evil Coordinate
題目
連結：https://ac.nowcoder.com/acm/contest/10272/E
來源：牛客網
題目描述
A robot is standing on an infinite 2-dimensional plane. Programmed with a string \(s_1s_2\cdots s_n\)
of length n, where \(s_i \in \{\text{'U'}, \text{'D'}, \text{'L'}, \text{'R'}\}\) the robot will start moving from \((0, 0)(0,0)\) and will follow the instructions represented by the characters in the string.
More formally, let (x, y)(x,y) be the current coordinate of the robot. Starting from (0, 0)(0,0), the robot repeats the following procedure nn times. During the ii-th time:
If \(s_i = \text{'U'}\) the robot moves from (x, y)(x,y) to (x, y+1)(x,y+1);
If \(s_i = \text{'D'}\)the robot moves from (x, y)(x,y) to (x, y-1)(x,y−1);
If \(s_i = \text{'L'}\)
If \(s_i = \text{'R'}\)the robot moves from (x, y)(x,y) to (x+1, y)(x+1,y).
However, there is a mine buried under the coordinate \((m_x, m_y)\). If the robot steps onto \((m_x, m_y)\)during its movement, it will be blown up into pieces. Poor robot!
Your task is to rearrange the characters in the string in any order, so that the robot will not step onto \((m_x, m_y)\)
輸入描述:
There are multiple test cases. The first line of the input contains an integer TT indicating the number of test cases. For each test case:
The first line contains two integers \(m_x\) and \(m_y\)
\((-10^9 \le m_x, m_y \le 10^9)\)indicating the coordinate of the mine.
The second line contains a string \(s_1s_2\cdots s_n\)of length n \((1 \le n \le 10^5)\)
, \(s_i \in \{\text{'U'}, \text{'D'}, \text{'L'}, \text{'R'}\}\) indicating the string programmed into the robot.
It's guaranteed that the sum of nn of all test cases will not exceed \(10^6\)
輸出描述:
For each test case output one line. If a valid answer exists print the rearranged string, otherwise print "Impossible" (without quotes) instead. If there are multiple valid answers you can print any of them.
示例1
輸入
5
1 1
RURULLD
0 5
UUU
0 3
UUU
0 2
UUU
0 0
UUU
輸出
LDLRUUR
UUU
Impossible
Impossible
Impossible
題意
給你一個x，y點和一個由UDLR組成的字串，機器人從（0，0）位置開始走。
重排這個字串使得機器人在走的過程中不會經過x,y
思路
x=0 && y=0的情況顯然不合法。
先把起點和重點之間的路徑抽象化成一個矩形，這樣如果每一次走都把當前方向的步數走完的話到達終點就有24種情況。列舉\(4!\) 種情況 如果走到了雷上就返回false 如果有合法的方案就輸出，否則的話就輸出Impossible
剛開始我想自己手動列舉所有情況，寫了快300行只過了三分之二的資料，於是考慮用next_permutation函式自動列舉每種情況。
程式碼(AC)
#include <bits/stdc++.h>
using namespace std;
string ans = "";
int x,y;
bool isok(char now[],int u,int d,int l,int r)
{
int cnt = 0;
int xx = 0,yy = 0;
ans = "";
for(int i=0; i<4; i++)
{
if(now[i] == 'U')
{
while(u--)
{
ans += 'U';
yy ++;
if(xx == x && yy == y) return false;
}
}
else if(now[i] == 'D')
{
while(d--)
{
ans += 'D';
yy --;
if(xx == x && yy == y) return false;
}
}
else if(now[i] == 'L')
{
while(l--)
{
ans += 'L';
xx --;
if(xx == x && yy == y) return false;
}
}
else if(now[i] == 'R')
{
while(r--)
{
ans += 'R';
xx ++ ;
if(xx == x && yy == y) return false;
}
}
}
return true;
}
void Solve()
{
scanf("%d%d",&x,&y);
string s;
cin>>s;
if(x==0 && y==0)
{
printf("Impossible");
return;
}
int len = s.size();
int xx = 0,yy = 0;
int u=0,d=0,l=0,r=0;
for(int i=0; i<len; i++)
{
if(s[i] == 'U') u++;
else if(s[i] == 'D') d++;
else if(s[i] == 'L') l++;
else r++;
}
int f = 0;
char ch[] = {'U','D','L','R'};
for(int i=0; i<24; i++)
{
if(isok(ch,u,d,l,r))
{
cout<<ans;
f = 1;
break;
}
next_permutation(ch,ch+4);
}
if(!f) cout<<"Impossible";
}
int main()
{
int t;
cin>>t;
while(t--)
{
Solve();
if(t) puts("");
}
}
程式碼(首次嘗試)
#include <bits/stdc++.h>
using namespace std;
void Solve()
{
int x,y;
scanf("%d%d",&x,&y);
string s;
cin>>s;
if(x==0 && y==0)
{
printf("Impossible");
return;
}
int len = s.size();
int xx = 0,yy = 0 , f = 1;
int u=0,d=0,l=0,r=0;
int ok = 0;
for(int i=0; i<len; i++)
{
if(s[i] == 'U') yy++ , u++;
else if(s[i] == 'D') yy-- , d++;
else if(s[i] == 'L') xx-- , l++;
else xx++ , r++;
if(xx == x && yy == y)
f = 0;
}
if(f) cout<<s;
else if(xx == x && yy == y) printf("Impossible");
else if(l==0 && r==0)
{
if(x !=0 )
{
while(u--) printf("U");
while(d--) printf("D");
}
else
{
if(y>0)
{
if(u-d>=y)
printf("Impossible");
else
{
while(d--) printf("D");
while(u--) printf("U");
}
}
else
{
if(d-u>=abs(y))
printf("Impossible");
else
{
while(u--) printf("U");
while(d--) printf("D");
}
}
}
}
else if(u==0 && d==0)
{
if(y !=0 )
{
while(l--) printf("L");
while(r--) printf("R");
}
else
{
if(x>0)
{
if(r-l>=x)
printf("Impossible");
else
{
while(l--) printf("L");
while(r--) printf("R");
}
}
else
{
if(l-r>=abs(y))
printf("Impossible");
else
{
while(r--) printf("R");
while(l--) printf("L");
}
}
}
}
else // 三個以上不是0
{
int now_x = 0,now_y = 0;
int uu = u,dd = d,ll = l,rr = r;
int f = 0;
while(uu--)
{
now_y++;
if(now_x==x && now_y==y)
{
f = 1;
now_y--;
uu++;
break;
}
}
if(f) // 向上走遇到雷
{
if(xx!=0) //如果左右偏移不是0
{
ok = 1;
while(l--) printf("L");
while(r--) printf("R");
while(u--) printf("U");
while(d--) printf("D");
}
else // 如果左右偏移是0
{
if(yy!=y)
{
ok = 1;
while(l--) printf("L");
while(u--) printf("U");
while(d--) printf("D");
while(r--) printf("R");
}
else
{
printf("Impossible");
}
}
}
else // 向上走沒遇到雷
{
while(dd--) // 向上走再向下走
{
now_y--;
if(now_x==x && now_y==y)
{
f = 1;
now_y++;
dd++;
break;
}
}
if(f) // 向上走再向下走遇到雷
{
if(xx!=0) // 橫向偏移不是0 直接走
{
ok = 1;
while(l--) printf("L");
while(r--) printf("R");
while(u--) printf("U");
while(d--) printf("D");
}
else // 橫向偏移是0 考慮U和
{
if(yy!=y)
{
ok = 1;
while(l--) printf("L");
while(u--) printf("U");
while(d--) printf("D");
while(r--) printf("R");
}
else
{
printf("Impossible");
}
}
}
else // 向上走向下走沒有遇到雷-------------------------------------------------------------------
{
// 先左走
now_x = 0,now_y = 0;
int uu = u,dd = d,ll = l,rr = r;
int f = 0;
while(ll--)
{
now_x--;
if(now_x==x && now_y==y)
{
f = 1;
now_x++;
ll++;
break;
}
}
if(f) // 向左走遇到雷
{
if(yy!=0) //如果上下偏移不是0
{
ok = 1;
while(u--) printf("U");
while(d--) printf("D");
while(l--) printf("L");
while(r--) printf("R");
}
else // 如果上下偏移是0
{
if(xx!=x)
{
ok = 1;
while(u--) printf("U");
while(l--) printf("L");
while(r--) printf("R");
while(d--) printf("D");
}
else
{
printf("Impossible");
}
}
}
else // 往左走沒遇到雷
{
while(rr--) // 往左走再往右走
{
now_x++;
if(now_x==x && now_y==y)
{
f = 1;
now_y--;
rr++;
break;
}
}
if(f) // 往左走再往右走
{
if(yy!=0) // 縱向偏移不是0 直接走
{
ok = 1;
while(u--) printf("U");
while(d--) printf("D");
while(l--) printf("L");
while(r--) printf("R");
}
else // 縱向偏移是0
{
if(xx!=x)
{
while(u--) printf("U");
while(l--) printf("L");
while(r--) printf("R");
while(d--) printf("D");
}
else
{
printf("Impossible");
}
}
}
else
{
while(u--) printf("U");
while(d--) printf("D");
while(l--) printf("L");
while(r--) printf("R");
}
}
}
}
}
}
int main()
{
int t;
cin>>t;
while(t--)
{
Solve();
if(t) puts("");
}
}