Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Karell Incorporated has designed a new exploration robot that has the ability to explore new terrains, this new robot can move in all kinds of terrain, it only needs more fuel to move in rough terrains, and less fuel in plain terrains. The only problem this robot has is that it can only move orthogonally, the robot can only move to the grids that are at the North, East, South or West of its position.

The Karell`s robot can communicate to a satellite dish to have a picture of the terrain that is going to explore, so it can select the best route to the ending point, The robot always choose the path that needs the minimum fuel to complete its exploration, however the scientist that are experimenting with the robot, need a program that computes the path that would need the minimum amount of fuel. The maximum amount of fuel that the robot can handle is 9999 units

The Terrain that the robot receives from the satellite is divided into a grid, where each cell of the grid is assigned to the amount of fuel the robot would need to pass thought that cell. The robot also receives the starting and ending coordinates of the exploration area.

Input

The first line of the input file is the number of tests that must be examined.

The first line of the test is the number of rows and columns that the grid will contain. The rows and columns will be 0 < row100 , 0 < column100

The next lines are the data of the terrain grid

The last line of the test has the starting and ending coordinates.

Output

One line, for each test will have the amount of fuel needed by the robot

Sample Input

3
5 5
1 1 5 3 2
4 1 4 2 6
3 1 1 3 3 
5 2 3 1 2
2 1 1 1 1
1 1 5 5 
5 4
2 2 15 1
5 1 15 1
5 3 10 1
5 2 1 1 
8 13 2 15
1 1 1 4 
10 10
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 10 10

Sample Output

10
15
19

解析:STL中优先级队列与普通的队列不太一样,在于取出队列的元素不是从队列的首部取出,而是根据设定的元素优先级来取出对象.这一道题中,很显然是一道最短路径算法的题目,很容易想到使用Dijkstra算法,将临近的点加入到队列中,每次取出代价最小的元素,若该元素已经被访问过,则出列;否则将其临近的点加入到队列中,知道到达终点.

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;

int dir_x[] = {-1, 0, 1, 0};    // 上下左右四个方向
int dir_y[] = {0, 1, 0, -1};

struct Node {
	int x;
	int y;
	int cost;
	Node(int a, int b, int c) {
		x = a;
		y = b;
		cost = c;
	}
	friend bool operator < (Node a, Node b) {
		return a.cost > b.cost;    // 返回">"的结果使得队列默认为最小堆
	}
};

int main() {
	int t;
	int city[101][101];
	bool visit[101][101];          // 记录节点是否已经访问
	cin >> t;
	while (t--) {
		int m, n;
		cin >> m >> n;
		memset(visit, false, sizeof(visit));
		for (int i = 1; i <= m; i++) {
			for (int j = 1; j <= n; j++) {
				cin >> city[i][j];
			}
		}
		int s_x, s_y, e_x, e_y;     // 记录起始坐标和终点坐标
		cin >> s_x >> s_y >> e_x >> e_y;
		visit[s_x][s_y] = true;
		priority_queue<Node> q;
		q.push(Node(s_x, s_y, city[s_x][s_y]));
		Node top = q.top();         // 取出头顶元素
		while (!visit[e_x][e_y]) {
			for (int i = 0; i < 4; i++) {
				int x = top.x + dir_x[i];
				int y = top.y + dir_y[i];
				if (x >= 1 && x <= m && y >= 1 && y <= n && !visit[x][y]) {
					q.push(Node(x, y, top.cost + city[x][y]));
				}
			}
			top = q.top();          // 若头顶元素已经访问过,则使其出队
			while (visit[top.x][top.y]) {
				q.pop();
				top = q.top();;
			}
			visit[top.x][top.y] = true;
		}
		cout << top.cost << endl;
	}
	return 0;
}