广搜简单题

Catch That Cow

题目传送：POJ - 3278 - Catch That Cow

题解：点击即传送

迷宫问题

题目传送：POJ - 3984 - 迷宫问题

DFS也可以，见另一个题解

AC代码（BFS）：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};

int front, rear;
struct node {
    int x, y, pre;
}q[105];

int mp[5][5];
int vis[5][5];

void print(int x) {
    if(q[x].pre != -1) {
        print(q[x].pre);
        printf("(%d, %d)\n", q[x].x, q[x].y);
    }
}

void bfs(int x, int y) {
    front = 0;
    rear = 1;
    q[front].x = x;
    q[front].y = y;
    q[front].pre = -1;
    vis[x][y] = 1;
    while(front < rear) {
        for(int i = 0;  i < 4; i ++) {
            int xx = q[front].x + dx[i];
            int yy = q[front].y + dy[i];
            if(xx >= 0 && xx < 5 && yy >= 0 && yy < 5 && mp[xx][yy] != 1 && !vis[xx][yy]) {
                vis[xx][yy] = 1;
                q[rear].x = xx;
                q[rear].y = yy;
                q[rear].pre = front;
                rear ++;
            }
            if(xx == 4 && yy == 4) {
                print(front);
                return;
            }
        }
        front ++;//出队
    }
}

int main() {
    memset(vis, 0, sizeof(vis));
    for(int i = 0;  i< 5; i ++) {
        for(int j = 0; j < 5; j ++) {
            scanf("%d", &mp[i][j]);
        }
    }

    printf("(0, 0)\n");
    bfs(0, 0);
    printf("(4, 4)\n");

    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。