POJ 2243 广搜

来源：互联网发布：c语言关机恶搞程序编辑：程序博客网时间：2024/04/30 18:53

经典跳马问题。

题意：有一只国际象棋的马，给出马的起点终点坐标，求出马需要移动的最小次数。

广搜吧，没什么好说的。

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <vector>#include <stack>#include <queue>#include <cstring>#include <ctype.h>#define MAXN 200using namespace std;#define INF 2100000000int n, m;int map[MAXN][MAXN];int cnt[MAXN];int path[8][2] = {{-2, -1}, {-2, 1}, {-1, 2}, {1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}};bool vis[MAXN*MAXN];bool islegal(int row, int col){    if(row < 0 || row >= 8 || col < 0 || col >= 8 || vis[row*8+col]) return 0;    return 1;}void bfs(int sta, int end){    queue<int> q;    q.push(sta);    vis[sta] = 1;    while(!q.empty())    {        int fr = q.front();        q.pop();        if(fr == end) return;        for(int i = 0; i < 8; i++)        {            int row = fr/8 + path[i][1];            int col = fr%8 + path[i][0];            if(islegal(row, col))            {                int tt = row*8+col;                vis[tt] = 1;                q.push(tt);                cnt[tt] = cnt[fr] + 1;            }        }    }}int main(){    //freopen("C:/Users/zts/Desktop/in.txt", "r", stdin);    string a, b;    while(cin >> a >> b)    {        memset(vis, 0, sizeof(vis));        memset(cnt, 0, sizeof(cnt));        int s = (a[0]-'a')+(a[1]-'0'-1)*8;        int e = (b[0]-'a')+(b[1]-'0'-1)*8;        bfs(s, e);        //printf("To get from e2 to e4 takes 2 knight moves.")        cout << "To get from " << a << " to " << b << " takes " << cnt[e] << " knight moves." << endl;        //cout << cnt[e] << endl;    }    return 0;}

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