版权声明:希望能帮到弱校的ACMer成长,因为自己是弱校菜鸡~~~~ https://blog.csdn.net/Mr__Charles/article/details/82190463
HDU T1372 Knight Moves
题解:
题目意思就是给你起点和终点,让你以国际象棋的骑士的走法走到终点,其实就相当于玩中国象棋,马去吃掉其他子类似的走法。搞清楚八个方向怎么走,这道题就差不多了。
走法:
(辛苦大佬花时间做图,小弟借用,取自秦石秦草)
还有注意的一点是看清楚题呀,鄙人以为a-z,把判边界的判断都搞错了,实在是菜......
代码:
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
int sx,sy,ex,ey;
char s[3],e[3];
bool vis[8][8];
int mov[8][2] = {-2,-1,-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2}; //八个方向
struct Node{
int x,y,st;
}now,nex;
bool charge(int x,int y){
if(x < 0 || x >= 8 || y < 0||y >= 8||vis[x][y] == true)
return false;
return true;
}
int Bfs(int st){
now.x = sx;
now.y = sy;
now.st = st;
queue<Node> Q;
Q.push(now);
while(!Q.empty()){
now = Q.front();
Q.pop();
if(now.x == ex && now.y == ey)
return now.st;
for(int i = 0; i < 8; ++i){
nex.x = now.x + mov[i][0];
nex.y = now.y + mov[i][1];
if(charge(nex.x,nex.y)){
nex.st = now.st + 1;
vis[nex.x][nex.y] = true;
Q.push(nex);
}
}
}
}
int main(){
while(~scanf("%s%s",s,e)){ //用%s比较方便,不用处理换行和空格。
memset(vis,false,sizeof(vis));
sx = s[1] - '1';
sy = s[0] - 'a';
ex = e[1] - '1';
ey = e[0] - 'a';
printf("To get from %s to %s takes %d knight moves.\n",s,e,Bfs(0));
}
return 0;
}