Title Link: http://poj.org/problem?id=3050
Surface:
Hopscotch
Time Limit: 1000MS |
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Memory Limit: 65536K |
Total Submissions: 2563 |
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Accepted: 1823 |
Description
The cows play the child's game of hopscotch in a non-traditional. Instead of a linear set of numbered boxes into which to hop, the cows create a 5x5 rectilinear grid of digits parallel to The X and Y axes.
They then adroitly hop onto any digit in the grid and hop forward, backward, right, or left (never diagonally) to another Digit in the grid. They hop again (same rules) to a digit (potentially a digit already visited).
With a total of five intra-grid hops, their hops create a six-digit integer (which might has leading zeroes like 000201).
Determine the count of the number of distinct integers that can is created in this manner.
Input
* Lines 1..5:the grid, five integers per line
Output
* Line 1:the number of distinct integers so can be constructed
Sample Input
1 1 1 1 11 1 1 1 11 1 1 1 11 1 1 2 11 1 1 1 1
Sample Output
15
Hint
OUTPUT DETAILS:
111111, 111112, 111121, 111211, 111212, 112111, 112121, 121111, 121112, 121211, 121212, 211111, 211121, 212111, and 212121 can be constructed. No other values is possible.
Source
Usaco 2005 November Bronze
Main topic:
A cow in the 5*5 can start jumping from any point, each can jump in 4 directions, Hop 5 times, according to walk through the square to form a length of 6 sequence, asked how many of these sequences? (squares can be skipped repeatedly.) )
Solving:
The topic hangs when writes is the exhaustion search, sees the picture so small, estimates is the violent search means. The enumeration starts at each point, jumps to 5 steps, stops, and records the path that is formed at this point, using the set row weight count.
Code:
#include <iostream> #include <cstdio> #include <string> #include <set> #include <vector> Using namespace Std;char map[6][6],path[10];//Four directions int dir[4][2]={0,1,-1,0,0,-1,1,0};//count set <string> cnt; String tmp;//determines if the square inside bool inside (int x,int y) {if (x>=1&&x<=5&&y>=1&&y<=5) return True;else return false;} Deep search void dfs (int x,int y,int step) {int tx,ty;path[step]=map[x][y];if (step==5) {//record path tmp= ""; for (int i=0;i<=5;i++) tmp+=path[i]; Cnt.insert (TMP); return;} Jump in four directions for (int i=0;i<4;i++) {tx=x+dir[i][0];ty=y+dir[i][1];if (inside (tx,ty)) DFS (tx,ty,step+1);}} int main () {//read in for (int i=1;i<=5;i++) for (int j=1;j<=5;j++) scanf ("%c", &map[i][j]);//enumeration starts at each point for (int i =1;i<=5;i++) for (int j=1;j<=5;j++) DFS (i,j,0);//Output printf ("%d\n", Cnt.size ()); return 0;}
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POJ 3050 Hopscotch (exhaustion search)