VK Cup 2017

传送门

Examples

input

2 3 2.**X..

output

RL

input

5 6 14..***.*...X...*.....*.**....*.

output

DLDDLLLRRRUURU

input

3 3 4****X****

output

IMPOSSIBLE

题意：

要求走出一个圈出来，使得走的路径是字典序最小的，一共要走K步，其中‘*’表示不能走，‘.’表示能走；

方法：先跑一边最短路看是否能满足条件，不行就IMPOSSIBLE，如果可以，那么就走一边BFS就完了，虽说要求字典序最小，让方向按字典序排序就好了

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.///             __.'              ~.   .~              `.__///           .'//                  \./                  \\`.///        .'//                     |                     \\`.///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.///     .'//.-"                 `-.  |  .-'                 "-.\\`.///   .'//______.============-..   \ | /   ..-============.______\\`./// .'______________________________\|/______________________________`.#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=1e19+10;//const int dx[]={-1,0,1,0,1,-1,-1,1};//const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e3+10;const int maxx=1e6+10;const double EPS=1e-10;const double eps=1e-10;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;/*void readString(string &s){static char str[maxx];scanf("%s", str);s = str;}*/int n,m,k;struct node{    int x,y,step;};int dx[]={1,0,0,-1};int dy[]={0,-1,1,0};char str[]={'D','L','R','U'};int d[1005][1005];int stx,sty;char s[maxn][maxn];queue<node>Q;int check(int x,int y){    if(s[x][y]=='.'||s[x][y]=='X')        return 1;    return 0;}string ans;void bfs(){    for(int i=0;i<=1000;i++)        for(int j=0;j<=1000;j++)            d[i][j]=INF;    d[stx][sty]=0;    Q.push((node){stx,sty,0});    int cnt=0;    W(!Q.empty())    {        cnt++;        node t=Q.front();        Q.pop();        for(int i=0;i<4;i++)        {            int xx=t.x+dx[i],yy=t.y+dy[i];            if(check(xx,yy))            {                if(d[xx][yy]>d[t.x][t.y]+1)                {                    Q.push((node){xx,yy,t.step+1});                    d[xx][yy]=d[t.x][t.y]+1;                }            }        }    }    if(cnt==1) puts("IMPOSSIBLE");    else    {        int dis=k;        for(int i=0;i<k;i++)        {            int pos=-1;            for(int j=0;j<4;j++)            {                int xx=stx+dx[j],yy=sty+dy[j];                if(check(xx,yy)&&d[xx][yy]<dis)                {                    pos=j;                    break;                }            }            dis--;            if(pos==-1)            {                puts("IMPOSSIBLE");return ;            }            stx=stx+dx[pos],sty=sty+dy[pos];            ans+=str[pos];        }        cout<<ans<<endl;    }    return ;}void solve(){    W(s_3(n,m,k)!=EOF)    {        ans.clear();        FOR(1,n,i)            scanf("%s",s[i]+1);        FOR(1,n,i)        {            FOR(1,m,j)            {                if(s[i][j]=='X')                    stx=i,sty=j;            }        }        bfs();    }}int main(){    //freopen( "in.txt" , "r" , stdin );    //freopen( "data.txt" , "w" , stdout );    int t=1;    //init();    //s_1(t);    for(int cas=1;cas<=t;cas++)    {        //printf("Case #%d: ",cas);        solve();    }}