【Codeforces 538 G】Berserk Robot

传送门

好像没用什么算法。。。

考虑把坐标$(x,y)\to (y+x,y-x)$
然后每一步位移变成$(\underline{+}1,\underline{+}1)$
再每步$x,y$都额外$+1$后除以2
那每个信息的$(x,y)\to ((x+y+t)/2,(y-x+t)/2)$

这样$x,y$之间没有关联了
考虑对于$x$
现在每一步要么不变要么$+1$
令$k$为每一次总循环的位移，$w=t\%l,v=\lfloor \frac{t}{l}\rfloor$
按照$w$排序后对于前后两个$i,j=i+1$
合法就必须满足$x_j-x_i\le k+\Delta v\ast k,x_j-x_i\ge \Delta v\ast k$
移项后就得一个到$k$的大小限制

得到所有的限制后取一个合法的$k$再还原即可

#include
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline ll readll(){
    char ch=gc();
    ll res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	return top;
}
template<typename tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<typename tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int N=200005,M=2000005;
int n,l,vx[M],vy[M];
struct node{
	ll x,y,t,w,v;
	node(ll a=0,ll b=0,ll c=0,ll d=0,ll e=0):x(a),y(b),t(c),w(d),v(e){}
	friend inline bool operator <(cs node &a,cs node &b){
		return a.w<b.w;
	}
}a[N];
inline void get(int *ans){
	ll L=0,R=l;
	for(int j=2;j<=n;j++){
		ll dv=a[j].v-a[j-1].v,dw=a[j].w-a[j-1].w,dx=a[j].x-a[j-1].x;
		if(dv==0){
			if(dx<0||dx-dw>0){puts("NO");exit(0);}
		}
		if(dv>0){
			chemx(L,(ll)ceil(1.0*(dx-dw)/dv)),chemn(R,(ll)floor(1.0*dx/dv));
		}
		if(dv<0){
			chemx(L,(ll)ceil(1.0*dx/dv)),chemn(R,(ll)floor(1.0*(dx-dw)/dv));
		}
	}
	if(L>R){puts("NO");exit(0);}
	int k=L;
	for(int i=2;i<=n;i++){
		int dx=(a[i].x-a[i].v*k)-(a[i-1].x-a[i-1].v*k),del=a[i].w-a[i-1].w;
		if(dx<0){puts("NO");exit(0);}
		for(int j=1;j<=del;j++){
			if(j<=dx)ans[j+a[i-1].w]=1;
			else ans[j+a[i-1].w]=0;
		}
	}
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),l=read();
	for(int i=1;i<=n;i++){
		ll t=readll(),x=readll(),y=readll();
		if((x+y+t)&1)return puts("NO"),0;
		if((x-y+t)&1)return puts("NO"),0;
		a[i].x=(x+y+t)>>1,a[i].y=(y-x+t)>>1,a[i].t=t,a[i].w=t%l,a[i].v=t/l;
	};
	n++,a[n]=node(0,0,0,0,0),n++,a[n]=node(0,0,0,l,-1);
	sort(a+1,a+n+1);
	get(vx);
	for(int i=1;i<=n;i++)swap(a[i].x,a[i].y);
	get(vy);
	for(int i=1;i<=l;i++){
		if(vx[i]==0&&vy[i]==0)putchar('D');
		if(vx[i]==0&&vy[i]==1)putchar('L');
		if(vx[i]==1&&vy[i]==0)putchar('R');
		if(vx[i]==1&&vy[i]==1)putchar('U');
	}return 0;
}