Openjudge 仙岛求药 BFS queue

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题目链接：http://noi.openjudge.cn/ch0205/2727/

是一个简单的BFS，这里重新熟悉了一下queue的使用

头文件#include<queue>

成员函数：

back

returns a reference to last element of a queue

empty

true if the queue has no elements

front

returns a reference to the first element of a queue

pop

removes the top element of a queue

push

adds an element to the end of the queue

size

returns the number of items in the queue

一般而言定义的时候这样 queue<int> q;

取出首个元素的操作 q.front() 弹出首个元素无返回值的操作 q.pop()

注意queue的清空没有clear这样的操作，但是可以用while不停的pop掉

写这么简单的题的时候遇到了一个比较蛋疼的bug,首先是上下左右，即dx, dy数组写的时候脑袋抽了

然后是多个案例的处理，我在完成第一个后，没有清空变量，即第二个案例没有很好的初始化

代码如下

// ConsoleApplication1.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include<iostream>#include<stdio.h>#include<queue>#include<memory.h>using namespace std;struct pos{int x, y, len;};queue<pos> q;int nx, ny, start_x, start_y, des_x, des_y;char ch;int con[25][25];int dx[4] = { 1,-1,0,0 };int dy[4] = { 0,0,-1,1 };bool visit[25][25];int bfs(int x, int y){pos t1, t2;t1.x = x; t1.y = y; t1.len = 0;visit[x][y] = false;//cout << x << " " << y << endl;q.push(t1);while (!q.empty()){t1 = q.front();q.pop();for (int i = 0; i < 4; i++) {t2.x = t1.x + dx[i]; t2.y = t1.y + dy[i];t2.len = t1.len + 1;if (t2.x == des_x && t2.y == des_y) return t2.len;if (t2.x < 0 || t2.x >= nx || t2.y < 0 || t2.y >= ny) continue;if (visit[t2.x][t2.y] == false) continue;if (con[t2.x][t2.y] == 2) continue;visit[t2.x][t2.y] = false;q.push(t2);//cout << t2.x << " "<<t2.y << endl;}}return -1;}int main(){while (1) {memset(visit, true, sizeof(visit));memset(con, 0, sizeof(con));while (!q.empty()) q.pop();cin >> nx >> ny;if (nx == 0 && ny == 0) break;for (int i=0; i<nx; i++)for (int j = 0; j < ny; j++){cin >> ch;if (ch == '.') con[i][j] = 1;if (ch == '#') con[i][j] = 2;if (ch == '@'){con[i][j] = 3;start_x = i; start_y = j;}if (ch == '*') {con[i][j] = 4;des_x = i; des_y = j;}}cout<<bfs(start_x, start_y)<<endl;/*for (int i = 0; i < nx; i++) {for (int j = 0; j < ny; j++)if (visit[i][j] == false) cout << " 1";else cout << " 0";cout << endl;}*/}    return 0;}

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