rt。

题目就是一道基本的广搜模版。

题意简述：
有一个 $n \times n$ 的地图，从 $mp_{x1,y1}$ 走到 $mp_{x2,y2}$，只能上下左右走，每次移动一次需要 $m$ 的时间。$1 \le n \le 1000, 1 \le m \le 10000000000$。

#include <bits/stdc++.h>
#define ll long long
#define int unsigned ll
using namespace std;
struct Node {
	int x, y; // 当前坐标 
	int step; // 当前走的步数 
};
int n, m;
// int mp[1005][1005];
int qx, qy, zx, zy;
queue <Node> q;
bool vis[10005][10005];
signed main() {
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
	cin >> n >> m;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j) {
			cin >> vis[i][j];
		}
	cin >> qx >> qy >> zx >> zy;
	Node u;
	u.x = qx;
	u.y = qy;
	u.step = 0;
	q.push(u);
	vis[u.x][u.y] = 1;
	if (u.x == zx && u.y == zy) {
		cout << 0;
		return 0;
	}
	while (!q.empty()) {
		Node v;
		u = q.front();
		q.pop();
		v.x = u.x + 1;
		v.y = u.y;
		v.step = u.step + m;
		if (v.x <= n && !vis[v.x][v.y]) {
			vis[v.x][v.y] = 1;
			q.push(v);
		}
		if (v.x == zx && v.y == zy) {
			cout << v.step << endl;
			return 0;
		}
//		cout << v.x << " "<< v.y << " " << v.step << endl;
//------------------1------------------------
		v.x = u.x - 1;
		v.y = u.y;
		v.step = u.step + m;
		if (v.x > n && !vis[v.x][v.y]) {
			vis[v.x][v.y] = 1;
			q.push(v);
		}
		if (v.x == zx && v.y == zy) {
			cout << v.step << endl;
			return 0;
		}
//		cout << v.x << " "<< v.y << " " << v.step << endl;
//------------------2------------------------
		v.x = u.x;
		v.y = u.y + 1;
		v.step = u.step + m;
		if (v.x <= n && !vis[v.x][v.y]) {
			vis[v.x][v.y] = 1;
			q.push(v);
		}
		if (v.x == zx && v.y == zy) {
			cout << v.step << endl;
			return 0;
		}
//		cout << v.x << " "<< v.y << " " << v.step << endl;
//------------------3---------------------------
		v.x = u.x;
		v.y = u.y - 1;
		v.step = u.step + m;
		if (v.y > 0 && !vis[v.x][v.y]) {
			vis[v.x][v.y] = 1;
			q.push(v);
		}
		if (v.x == zx && v.y == zy) {
			cout << v.step << endl;
			return 0;
		}
//		cout << v.x << " "<< v.y << " " << v.step << endl;
//-----------------4-----------------
	}
}
/*
4 2
0 1 1 0
0 0 0 0
1 1 1 0
1 1 1 0
1 1 4 4
*/