poj 3050 Hopscotch

最新推荐文章于 2022-03-10 20:21:29 发布

superkunn

最新推荐文章于 2022-03-10 20:21:29 发布

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CC 4.0 BY-SA版权

文章标签： bfs

本文链接：https://blog.youkuaiyun.com/superkunn/article/details/79545005

本文探讨了一种基于5x5数字网格的奶牛跳跃游戏，通过特定的跳跃规则形成六位数整数。文章提供了完整的算法实现，利用广度优先搜索算法（BFS）来确定所有可能的不同整数，并通过哈希映射去除重复值。

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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

Sample Output

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.

#include <iostream>
#include <queue>
#include <map>
#include <cstdio>
#include <algorithm>
#include <string>

using namespace std;

int mapp[6][6];
int ans;

map<int,bool> mm;


struct node{
    int x;
    int y;
    int z;
    int s;
};

int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};


void bfs(int x,int y,int z){
    node nd;
    nd.x=x;
    nd.y=y;
    nd.z=z;
    nd.s=mapp[x][y];

    queue<node> q;
    q.push(nd);

    while(!q.empty()){
        nd=q.front();
        q.pop();

        if(nd.z>=6)break;

        for(int i=0;i<4;i++){
            int nx=nd.x+dx[i];
            int ny=nd.y+dy[i];

            if(1<=nx&&nx<=5&&1<=ny&&ny<=5){
                node t;
                t.x=nx;
                t.y=ny;
                t.z=nd.z+1;
                t.s=nd.s*10+mapp[t.x][t.y];

                if(t.z==5){
                    if(mm[t.s]!=1){
                        ans++;
                        mm[t.s]=1;
                    }
                }

                q.push(t);
            }
        }

    }
    return;
}

int main(){


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

    ans=0;
    for(int i=1;i<=5;i++){
        for(int j=1;j<=5;j++){
            bfs(i,j,0);
        }
    }
    printf("%d",ans);

    return 0;
}

map set都可以