题目传送门
You were dreaming that you are traveling to a planet named Planetforces on your personal spaceship. Unfortunately, its piloting system was corrupted and now you need to fix it in order to reach Planetforces.
Space can be represented as the XY plane. You are starting at point (0,0), and Planetforces is located in point (px,py).
The piloting system of your spaceship follows its list of orders which can be represented as a string s. The system reads s from left to right. Suppose you are at point (x,y) and current order is si:
if si=U, you move to (x,y+1);
if si=D, you move to (x,y−1);
if si=R, you move to (x+1,y);
if si=L, you move to (x−1,y).
Since string s could be corrupted, there is a possibility that you won't reach Planetforces in the end. Fortunately, you can delete some orders from s but you can't change their positions.
Can you delete several orders (possibly, zero) from s in such a way, that you'll reach Planetforces after the system processes all orders?
Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases.
Each test case consists of two lines. The first line in each test case contains two integers px and py (−105≤px,py≤105; (px,py)≠(0,0)) — the coordinates of Planetforces (px,py).
The second line contains the string s (1≤|s|≤105: |s| is the length of string s) — the list of orders.
It is guaranteed that the sum of |s| over all test cases does not exceed 105.
Output
For each test case, print "YES" if you can delete several orders (possibly, zero) from s in such a way, that you'll reach Planetforces. Otherwise, print "NO". You can print each letter in any case (upper or lower).
Example
input

6
10 5
RRRRRRRRRRUUUUU
1 1
UDDDRLLL
-3 -5
LDLDLDDDR
1 2
LLLLUU
3 -2
RDULRLLDR
-1 6
RUDURUUUUR

output

YES
YES
YES
NO
YES
NO

Note
In the first case, you don't need to modify s, since the given s will bring you to Planetforces.

In the second case, you can delete orders s2, s3, s4, s6, s7 and s8, so s becomes equal to "UR".

In the third test case, you have to delete order s9, otherwise, you won't finish in the position of Planetforces.

题目解析

我们发现，我们可以任意删除操作，所以我们只要留下朝终点的方向的操作即可，然后我们就可以判断个数是否能够到达终点即可。
代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#define maxn 1000039
using namespace std;
//#define debug
typedef int Type;
inline Type read(){
	Type sum=0;
	int flag=0;
	char c=getchar();
	while((c<'0'||c>'9')&&c!='-') c=getchar();
	if(c=='-') c=getchar(),flag=1;
	while('0'<=c&&c<='9'){
		sum=(sum<<1)+(sum<<3)+(c^48);
		c=getchar();
	}
	if(flag) return -sum;
	return sum;
}
int T;
int canx,cany;
int x,y;
int l,r,u,d,n;
char s[maxn];
int main(){
    cin>>T;
    while(T--){
    	cin>>x>>y;
    	l=r=u=d=0;
		cin>>s; n=strlen(s);
    	for(int i=0;i<n;i++)
    	    if(s[i]=='L') l++;
    	    else if(s[i]=='R') r++;
    	    else if(s[i]=='U') u++;
    	    else d++;
    	if(y>=0){
			if(y<=u) canx=1;
    	    else canx=0;
		}
    	if(y<0){
			if(-y<=d) canx=1;
    	    else canx=0;
		}
    	if(x>=0){
    		if(x<=r) cany=1;
    	    else cany=0;
		}
    	if(x<0){
    		if(-x<=l) cany=1;
    	    else cany=0;
		}
    	if(canx==1&&cany==1) printf("YES\n");
    	else printf("NO\n");
	}
	return 0;
}