POJ 3984 迷宫问题 bfs

题目链接：迷宫问题 

天啦撸。最近怎么了。小bug缠身，大bug 不断。然这是我大腿第一次给我dbug。虽然最后的结果是。我............bfs入队列的是now..............

然后。保存路径的一种用的string 。一种用的数组。大同小异。根据就是我bfs 先搜到的绝壁就是步数最少的、

附代码：

pre 数组

/*
 很简单的广搜。如果不是+路径输出的话。。
 保存路径。
 */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <string>
using namespace std;

struct Node {
    int x, y;
}now, temp, ans;

int mp[10][10];
int vis[10][10];
Node que[210];
Node pre[210];

int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};

bool check(Node a) {
    if (a.x >= 0 && a.x < 5 && a.y >= 0 && a.y < 5 && !vis[a.x][a.y] && mp[a.x][a.y] == 0)
        return true;
    return false;
}

int head = 0, tail = 0;

void bfs() {
    while(head < tail) {
        now = que[head++];
        if (now.x == 4 && now.y == 4) {
            return;
        }
        for (int i=0; i<4; ++i) {
            temp.x = now.x + dir[i][0];
            temp.y = now.y + dir[i][1];

            if (check(temp)) {
                que[tail++] = temp;
                vis[temp.x][temp.y] = 1;
                int id = temp.x * 5 + temp.y;
                pre[id].x = now.x, pre[id].y = now.y;
            }
        }
    }
}

void outPre(int num) {
    if (pre[num].x == -1 && pre[num].y == -1)
        return;
    else outPre(5*pre[num].x+pre[num].y);
    cout << "(" << pre[num].x << ", " << pre[num].y << ")
";
}

int main() {
    head = 0, tail = 0;
    memset(vis, 0, sizeof(vis));

    for (int i=0; i<5; ++i) {
        for (int j=0; j<5; ++j) {
            cin >> mp[i][j];
        }
    }
    now.x = 0, now.y = 0;
    vis[0][0] = 1;
    que[tail++] = now;
    pre[0].x = -1, pre[0].y = -1;
    bfs();

    outPre(24);
    cout << "(4, 4)
";
    return 0;
}

string 

/*
 很简单的广搜。如果不是+路径输出的话。想了一下 好像是dfs？好像不是啊。

 */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <string>
using namespace std;

struct Node {
    int x, y;
    string ansStep;
}now, temp, ans;

int mp[10][10];
int vis[10][10];
Node que[210];

int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
bool check(Node a) {
    if (a.x >= 0 && a.x < 5 && a.y >= 0 && a.y < 5 && !vis[a.x][a.y] && mp[a.x][a.y] == 0)
        return true;
    return false;
}

int head = 0, tail = 0;

void bfs() {
    while(head < tail) {
        now = que[head++];
        if (now.x == 4 && now.y == 4) {
            now.ansStep += '';
            ans.ansStep = now.ansStep;
            return;
        }
        for (int i=0; i<4; ++i) {
            temp.x = now.x + dir[i][0];
            temp.y = now.y + dir[i][1];

            if (check(temp)) {
                vis[temp.x][temp.y] = 1;
                temp.ansStep = now.ansStep;
                temp.ansStep += temp.x + '0';
                temp.ansStep += temp.y + '0';
                que[tail++] = temp;
            }
        }
    }
}

int main() {
    head = 0, tail = 0;
    memset(vis, 0, sizeof(vis));

    for (int i=0; i<5; ++i) {
        for (int j=0; j<5; ++j) {
            cin >> mp[i][j];
        }
    }
    now.x = 0, now.y = 0;
    now.ansStep += "00";

    vis[0][0] = 1;
    que[tail++] = now;
    bfs();

    string ansstr = ans.ansStep;
    int len = ansstr.length();
    for (int i=0; i<len && i+2<len; i+=2) {
        cout << "(" << ansstr[i] << ", " << ansstr[i+1] << ")" << endl;
    }
    return 0;
}