题意是：搜索骑士在棋盘上从起点到终点所走的最短步数，注意的是骑士不是一格一格走的，而是跨各走的；
思路: 就是广度优先搜索bfs，主要是找到骑士的走法
int D[8][2] = {{-2,-1},{-1,-2},{1,-2},{2,-1},{-2,1},{-1,2},{1,2},{2,1}};
poj1915代码……
#include<iostream>
#include<queue>
using namespace std; // 624k 641ms bfs搞定，但是有点慢。
typedef struct
{
int x,y;
int sum;
}Map;
//int map[301][301];
int visit[301][301];
int D[8][2] = {{-2,-1},{-1,-2},{1,-2},{2,-1},{-2,1},{-1,2},{1,2},{2,1}};
int n,l;
int sx,sy,ex,ey;
int Min,x1,y1,i;
int bfs()
{
queue<Map> Q;
Map a,b;
Min = 10000;
a.x = sx;
a.y = sy;
a.sum = 0;
Q.push(a);
visit[sx][sy] = 1;
while( !Q.empty())
{
a = Q.front();
Q.pop();
//if(a.x<0 || a.x>l || a.y<0 || a.y>l)
// continue;
if(a.x == ex && a.y == ey)
{
if(a.sum < Min)
Min = a.sum;
}
for(i=0; i<8; i++)
{
x1 = a.x + D[i][0];
y1 = a.y + D[i][1];
if(x1>=0 && y1>=0 && x1<l && y1<l && !visit[x1][y1])
{
b.x = x1;
b.y = y1;
b.sum = a.sum + 1;
Q.push(b);
visit[x1][y1] = 1;
}
}
}
return Min;
}
int main()
{
scanf("%d",&n);
while(n--)
{
memset(visit,0,sizeof(visit));
//memset(map,0,sizeof(map));
scanf("%d",&l);
scanf("%d%d",&sx,&sy);
scanf("%d%d",&ex,&ey);
printf("%d/n",bfs());
}
return 0;
}
poj2243代码……
这题不同之处是要转换下坐标；
#include<iostream>
#include<queue>
#include<cmath>
using namespace std; //176K 454MS bfs()搞定……
typedef struct
{
int x,y;
int sum;
}Map;
int visit[9][9];
int D[8][2] = {{-2,-1},{-1,-2},{1,-2},{2,-1},{-2,1},{-1,2},{1,2},{2,1}}; //这个题目主要的是骑士的走法问题！！！
int sx,sy,ex,ey;
int Min;
int bfs()
{
int x1,y1,i;
queue<Map> Q;
Map a,b;
Min = 10000;
a.x = sx;
a.y = sy;
a.sum = 0;
Q.push(a);
visit[sx][sy] = 1;
while( !Q.empty())
{
a = Q.front();
Q.pop();
if(a.x == ex && a.y == ey)
{
if(a.sum < Min)
Min = a.sum;
}
for(i=0; i<8; i++)
{
x1 = a.x + D[i][0];
y1 = a.y + D[i][1];
if(x1>0 && y1>0 && x1<=8 && y1<=8 && !visit[x1][y1])
{
b.x = x1;
b.y = y1;
b.sum = a.sum + 1;
Q.push(b);
visit[x1][y1] = 1;
}
}
}
return Min;
}
int main()
{
char begin[5],end[5];
while(scanf("%s %s",begin,end) != EOF)
{
memset(visit,0,sizeof(visit));
sx = abs(begin[0]-'a'-8); //转换为对应的坐标
sy = begin[1]-'0';
ex = abs(end[0]-'a'-8);
ey = end[1]-'0';
printf("To get from %s to %s takes %d knight moves./n",begin,end,bfs());
}
return 0;
}