UVA 439 Knight Moves

 

// 题意：输入标准国际象棋棋盘上的两个格子，求马最少需要多少步从起点跳到终点

BFS求最短路：

bfs并维护距离状态cnt， vis记录是否访问过

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
#include<queue>
using namespace std;
int r1, c1, r2, c2;
const int N=8;
int vis[N][N];

const int dr[] = {-2,-2,-1,-1, 1,1, 2,2};
const int dc[] = {-1, 1,-2, 2,-2,2,-1,1};


struct State {
    int r, c;
    int cnt;
    State(int r=0, int c=0, int cnt=0):r(r),c(c),cnt(cnt) {}
    bool operator == (const State& rhs)
    {
        return r==rhs.r && c==rhs.c;
    }
};

void tr(char* p, int &r, int &c)
{
    r=p[0]-'a';
    c=p[1]-'1';
}

int bfs()
{
    State f(r1, c1, 0), t(r2, c2);
    if(f==t) return 0;
    
    queue<State> Q;
    Q.push(f);
    vis[r1][c1]=1;

    while(!Q.empty())
    {
        State s=Q.front(); Q.pop();
        if(s==t)
            return s.cnt;
        for(int i=0;i<8;i++)
        {
            int nr, nc;
            nr=s.r+dr[i];
            nc=s.c+dc[i];
            if(nr>=0 && nr<8 && nc>=0 && nc<8 && !vis[nr][nc])
            {
                Q.push(State(nr, nc, s.cnt+1));
                vis[nr][nc]=1;
            }
        }
    }
    return -1;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("./uva439.in" , "r", stdin);
#endif

    char p1[3]; 
    char p2[3]; 
    while(scanf("%s%s", p1, p2)!=EOF)
    {
        tr(p1, r1, c1);
        tr(p2, r2, c2);
        memset(vis, 0, sizeof vis);
        int cnt=bfs();
        printf("To get from %s to %s takes %d knight moves.\n", p1, p2, cnt);
        
    }

    return 0;
}

 

lrj代码：

vis记录是否访问过， d记录各个格子的距离（这个要学习）

// UVa439 Knight Moves
// Rujia Liu
// 题意：输入标准国际象棋棋盘上的两个格子，求马最少需要多少步从起点跳到终点
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

struct State {
  int r, c;
  State(int r, int c):r(r),c(c) {}
};

const int dr[] = {-2,-2,-1,-1, 1,1, 2,2};
const int dc[] = {-1, 1,-2, 2,-2,2,-1,1};
int d[8][8], vis[8][8];
  
int bfs(int r1, int c1, int r2, int c2) {
  if(r1 == r2 && c1 == c2) return 0;
  queue<State> Q;
  d[r1][c1] = 0;
  vis[r1][c1] = 1; 
  Q.push(State(r1, c1));
  while(!Q.empty()) {
    State s = Q.front(); Q.pop(); 
    for(int i = 0; i < 8; i++) {
      int newr = s.r + dr[i];
      int newc = s.c + dc[i];
      if(newr >= 0 && newr < 8 && newc >= 0 && newc < 8 && !vis[newr][newc]) {
        Q.push(State(newr, newc));
        vis[newr][newc] = 1;
        d[newr][newc] = d[s.r][s.c] + 1;
        if(newr == r2 && newc == c2) return d[newr][newc];
      }
    }
  } 
  return -1; 
} 
    
int main() {
  char s[9], t[9];
  while(scanf("%s%s", s, t) == 2) {
    memset(vis, 0, sizeof(vis));
    int ans = bfs(s[0]-'a', s[1]-'1', t[0]-'a', t[1]-'1');
    printf("To get from %s to %s takes %d knight moves.\n", s, t, ans);
  }
  return 0;
}