POJ3050-穷竭搜索

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 7014	Accepted: 4510

Description

The cows play the child's game of hopscotch in a non-traditional way. 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 have leading zeroes like 000201).

Determine the count of the number of distinct integers that can be created in this manner.

Input

• Lines 1..5: The grid, five integers per line

Output

• Line 1: The number of distinct integers that can be constructed

Sample Input

1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 2 1
1 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 are possible.

Source

USACO 2005 November Bronze

题意：

选取一个点，然后从这个点往上下左右任意一个方向移动，重复五次这个步骤，得到六个数字，问有多少种组合。

注意：

• 格子可以重复跳

解法：

对每一个格子进行dfs，将每一次得到的数字用set存储，因为set不能出现重复的键，所以最后输出set里面元素的数量即可。

代码：

#include <iostream>
#include <stdio.h>
#include <set>

using namespace std;

int mmap[7][7];
int dire[4][2] = {0,1,0,-1,1,0,-1,0};
set<int> num;

void dfs(int x, int y,int ans, int d){

    // if(d == 6){
    //     num.insert(ans);
    //     return;
    // }

    for(int i = 0; i < 4; i++){
        int xx = x + dire[i][0];
        int yy = y + dire[i][1];

        if(xx >= 1 && xx <= 5 && yy >= 1 && yy <= 5){
            if(d + 1 != 7){
                dfs(xx, yy, ans * 10 + mmap[xx][yy], d+1);
            }else{
                num.insert(ans);
            }
            
        }
    }
}

int main(){

    for(int i = 1; i <= 5; i++){
        for(int j = 1; j <= 5; j++){
            scanf("%d",&mmap[i][j]);
        }
    }

    for(int i = 1; i <= 5; i++){
        for(int j = 1; j <= 5; j++){
            dfs(i,j,mmap[i][j],1);
        }
    }

    cout << num.size() << endl;
    // getchar();
    // getchar();

    return 0;
}