Knight Moves

题目描述

思路

代码描述

#include <cstdio>
#include <cstring>
#include <queue>

int n, m, ans;
struct Node {
	int x, y, z;
}st, ed, tmp;
std::queue<Node> q;
int mp[305][305];
bool vis[305][305];
int dirx[] = {0, -1, -2, -2, -1, 1, 2,  2,  1};
int diry[] = {0, -2, -1,  1,  2, 2, 1, -1, -2};
bool valid(int x, int y) {
	if (x < 0 || x >= m) return false;
	if (y < 0 || y >= m) return false;
	return true;
}
inline int read() {
	int s = 0, f = 1;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return s * f;	
}
int main() {
	n = read();
	while (n--) {
		m = read();
		st.x = read(), st.y = read(), st.z = 0;
		ed.x = read(), ed.y = read();
		tmp.x = st.x, tmp.y = st.y, tmp.z = st.z;
		while (!q.empty()) q.pop();
		memset(vis, 0, sizeof(vis));
		q.push(tmp);
		vis[tmp.x][tmp.y] = true;
		while (!q.empty()) {
			st = q.front();
			q.pop();
			if (st.x == ed.x && st.y == ed.y) {
				ans = st.z; 
				break;
			}
			for (int i = 1; i <= 8; ++i) {
				tmp.x = st.x + dirx[i];
				tmp.y = st.y + diry[i];
				tmp.z = st.z + 1;
				if (valid(tmp.x, tmp.y) && !vis[tmp.x][tmp.y]) {
					vis[tmp.x][tmp.y] = 1;
					q.push(tmp);
				}
			}
		}
		printf("%d
", ans);
	}
	return 0;
}
原文地址:https://www.cnblogs.com/liuzz-20180701/p/11598689.html