水。。。。。

直接bfs。。。

#include

#define INF 0x3f3f3f3f

using namespace std;

typedef long long ll;

inline int lowbit(int x){ return x & (-x); }

inline int read(){

int X = 0, w = 0; char ch = 0;

while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }

while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();

return w ? -X : X;

}

inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }

inline int lcm(int a, int b){ return a / gcd(a, b) * b; }

template

inline T max(T x, T y, T z){ return max(max(x, y), z); }

template

inline T min(T x, T y, T z){ return min(min(x, y), z); }

template

inline A fpow(A x, B p, C lyd){

A ans = 1;

for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;

return ans;

}

int n, m, f[155][155];

char g[155][155];

pair st, ed;

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

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

void init(){

for(int i = 0; i < m; i ++){

for(int j = 0; j < n; j ++){

if(g[i][j] == 'K')

st.first = i, st.second = j, g[i][j] = '.';

if(g[i][j] == 'H')

ed.first= i, ed.second = j, g[i][j] = '.';

}

}

}

bool inArea(int x, int y){

return x >= 0 && x < m && y >= 0 && y < n && g[x][y] == '.';

}

int bfs(){

memset(f, -1, sizeof f);

queue> q;

f[st.first][st.second] = 0;

q.push(st);

while(!q.empty()){

int x = q.front().first, y = q.front().second; q.pop();

for(int i = 0; i < 8; i ++){

int nx = x + dx[i];

int ny = y + dy[i];

if(inArea(nx, ny) && f[nx][ny] == -1){

f[nx][ny] = f[x][y] + 1;

if(nx == ed.first && ny == ed.second)

return f[nx][ny];

q.push(make_pair(nx, ny));

}

}

}

return -1;

}

int main(){

n = read(), m = read();

for(int i = 0; i < m; i ++) scanf("%s", g[i]);

init();

printf("%d\n", bfs());

return 0;

}

标签：return,进阶,int,st,nx,ny,&&,习题,CH2906

来源： https://www.cnblogs.com/onionQAQ/p/10542251.html