193 lines
5.5 KiB
Markdown
193 lines
5.5 KiB
Markdown
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# 137. 消息传输
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这道题目，普通广搜就可以解决。
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这里说一下几点注意事项：
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1、 题目描述中，注意 n 是列数，m是行数
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这是造成很多录友周赛的时候提交 返回 【运行错误】的罪魁祸首，如果 输入用例是 正方形，那没问题，如果后台输入用例是矩形， n 和 m 搞反了，就会数组越界。
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矩阵是 m * n ，但输入的顺序却是 先输入n 再输入 m。
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这会让很多人把矩阵的 n 和 m 搞反。
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其实规范出题，就应该是n 行，m列，然后 先输入n，在输入m。
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只能说 大厂出题的人，也不是专业出题的，所以会在 非算法方面一不小心留下很多 “bug”，消耗大家的精力。
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2、再写广搜的时候，可能担心会无限循环
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即 A 走到 B，B又走到A，A又走到B ，这种情况，一般来说 广搜都是用一个 visit数组来标记的。
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但本题不用，因为 不会重复走的，题图里的信号都是正数，根据距离判断大小 可以保证不走回头路。
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```CPP
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#include <iostream>
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#include <queue>
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#include <vector>
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using namespace std;
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const int inf = 1e6;
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int main () {
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int n, m, startx, starty;
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cin >> n >> m;
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cin >> startx >> starty;
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vector<vector<int>> grid(m, vector<int>(n));
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vector<vector<int>> dis(m, vector<int>(n, inf));
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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cin >> grid[i][j];
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}
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}
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queue<pair<int, int>> que;
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int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1};
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que.push(pair<int, int>(startx, starty));
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dis[startx][starty] = 0;
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while(!que.empty()) {
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pair<int, int> cur = que.front(); que.pop();
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for (int i = 0; i < 4; i++) {
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int newx = cur.first + dir[i][1];
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int newy = cur.second + dir[i][0];
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if (newx < 0 || newx >= m || newy < 0 || newy >= n || grid[cur.first][cur.second] == 0) continue;
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if (dis[newx][newy] > dis[cur.first][cur.second] + grid[cur.first][cur.second]) {
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dis[newx][newy] = dis[cur.first][cur.second] + grid[cur.first][cur.second];
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que.push(pair<int, int>(newx, newy));
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}
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}
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}
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int result = 0;
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (dis[i][j] == inf) {
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cout << -1 << endl;
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return 0;
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}
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result = max(result, dis[i][j]);
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}
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}
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cout << result << endl;
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}
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```
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## 其他语言版本
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### Java
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```Java
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import java.util.*;
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public class Main {
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static final int INF = 1000000;
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in);
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int n = scanner.nextInt();
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int m = scanner.nextInt();
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int startX = scanner.nextInt();
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int startY = scanner.nextInt();
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int[][] grid = new int[m][n];
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int[][] dis = new int[m][n];
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for (int i = 0; i < m; i++) {
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Arrays.fill(dis[i], INF);
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for (int j = 0; j < n; j++) {
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grid[i][j] = scanner.nextInt();
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}
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}
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Queue<int[]> queue = new LinkedList<>();
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int[][] directions = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
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queue.add(new int[]{startX, startY});
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dis[startX][startY] = 0;
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while (!queue.isEmpty()) {
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int[] current = queue.poll();
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for (int[] dir : directions) {
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int newX = current[0] + dir[0];
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int newY = current[1] + dir[1];
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if (newX >= 0 && newX < m && newY >= 0 && newY < n && grid[current[0]][current[1]] != 0) {
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if (dis[newX][newY] > dis[current[0]][current[1]] + grid[current[0]][current[1]]) {
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dis[newX][newY] = dis[current[0]][current[1]] + grid[current[0]][current[1]];
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queue.add(new int[]{newX, newY});
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}
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}
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}
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}
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int result = 0;
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (dis[i][j] == INF) {
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System.out.println(-1);
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return;
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}
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result = Math.max(result, dis[i][j]);
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}
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}
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System.out.println(result);
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scanner.close();
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}
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}
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```
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### Python
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```Python
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from collections import deque
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inf = 1000000
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def main():
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import sys
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input = sys.stdin.read
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data = input().split()
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index = 0
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n = int(data[index])
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m = int(data[index+1])
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startx = int(data[index+2])
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starty = int(data[index+3])
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index += 4
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grid = []
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dis = [[inf] * n for _ in range(m)]
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for i in range(m):
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grid.append([int(data[index+j]) for j in range(n)])
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index += n
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directions = [(0, 1), (1, 0), (-1, 0), (0, -1)]
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queue = deque()
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queue.append((startx, starty))
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dis[startx][starty] = 0
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while queue:
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curx, cury = queue.popleft()
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for dx, dy in directions:
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newx, newy = curx + dx, cury + dy
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if 0 <= newx < m and 0 <= newy < n and grid[curx][cury] != 0:
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if dis[newx][newy] > dis[curx][cury] + grid[curx][cury]:
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dis[newx][newy] = dis[curx][cury] + grid[curx][cury]
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queue.append((newx, newy))
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result = 0
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for i in range(m):
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for j in range(n):
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if dis[i][j] == inf:
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print(-1)
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return
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result = max(result, dis[i][j])
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print(result)
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if __name__ == "__main__":
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main()
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```
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