hdu1372Knight Moves

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1372

一个可以走8个方向。BFS求最短路

代码：

#include <cstdio>#include <cstring>#include <queue>using namespace std;int v[10][10];int stx,sty,enx,eny;struct node{    int x,y;    int step;};node ans;node t;int dx[] = {-1,-2,-2,-1,1,2,2,1};int dy[] = {-2,-1,1,2,2,1,-1,-2};void bfs(){    memset(v,0,sizeof(v));    queue<node> q;    node f;    f.x = stx;    f.y = sty;    f.step = 0;    q.push(f);    while(!q.empty())    {        node h = q.front();        q.pop();        if(h.x == enx && h.y == eny)        {            ans = h;            return ;        }        for(int i = 0 ;i < 8;++i)        {            int tx = h.x + dx[i];            int ty = h.y + dy[i];            if(tx >= 1 && tx <= 8 && ty >= 1 && ty <= 8 && !v[tx][ty])            {                t.x = tx;                t.y = ty;                t.step = h.step + 1;                q.push(t);                v[tx][ty] = 1;            }        }    }}int main(){    char s[5],s1[5];    while(~scanf("%s%s",s,s1))    {        stx = s[1] - '0';        sty = s[0] - 'a' + 1;        enx = s1[1] -'0';        eny = s1[0] - 'a' + 1;        if(stx == enx && sty == eny)        {            printf("To get from %s to %s takes 0 knight moves.\n",s,s1);            continue;        }        bfs();        printf("To get from %s to %s takes %d knight moves.\n",s,s1,ans.step);    }    return 0;}

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