Codevs 1215 Maze

time limit: 1 sspace limit: 128000 KBtitle level: Golden Gold Title Description Description

n integers are known to x1,x2,..., xn, and an integer k (k 3+7+12=22 3+7+19=29 7+12+19=38 3+12+19=34.
Now, ask you to figure out how many kinds of numbers are in total.
For example, in the above example, there is only one and a prime number: 3+7+19=29).

Enter a description input Description

Keyboard input, in the form of:
N, K (1<=n<=20,k X1,X2,...,XN (1<=xi<=5000000)

outputs description output Description

Screen output in the form of:
An integer (the number of species that satisfies the condition).

sample input to sample

4 3
3 7 12 19

Sample output Sample outputs

1

data size & Hint

(1<=n<=20,k (1<=xi<=5000000)

Portal point This expansion

The Simple BFS

#include <iostream>using namespacestd;intk,a,b,c,d,m;Charju[ -][ -];intfx[4]={0,0,-1,1},fy[4]={1,-1,0,0};BOOLssintAintb) {    intHead=0, tail=1, f[ -][2],x,y; f[tail][1]=A; f[tail][2]=b; JU[A][B]='#';  Do{Head++;  for(intI=0;i<4;++i) {x=f[head][1]+Fx[i]; Y=f[head][2]+Fy[i]; if(x>=0&&y>=0&&x<k&&y<k&&ju[x][y]=='.') {Tail++; f[tail][1]=x; f[tail][2]=y; Ju[x][y]='#'; }        }    } while(head<tail);}intMain () {intH=0; CIN>>m;  while(h<m) {h++; CIN>>K;  for(intI=0; i<k;++i) { for(intj=0; j<k;++j) {cin>>Ju[i][j]; if(ju[i][j]=='s') {a=i; b=J; JU[I][J]='.'; }                if(ju[i][j]=='e') {C=i; D=J; JU[I][J]='.';        }}} SS (A, B); if(ju[c][d]=='#') cout<<"YES"<<Endl; Elsecout<<"NO"<<Endl; }}

Codevs 1215 Maze