标签：要求 inpu tom push mes set eof main ||
迷宫问题
Time Limit: 1000MS
Memory Limit: 65536K
Description
定义一个二维数组：
int maze[5][5] = {
0, 1, 0, 0, 0,
0, 1, 0, 1, 0,
0, 0, 0, 0, 0,
0, 1, 1, 1, 0,
0, 0, 0, 1, 0,
};
Input
一个5 × 5的二维数组，表示一个迷宫。数据保证有唯一解。
Output
左上角到右下角的最短路径，格式如样例所示。
Sample Input
0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 1 0
Sample Output
(0, 0) (1, 0) (2, 0) (2, 1) (2, 2) (2, 3) (2, 4) (3, 4) (4, 4)
#include <iostream> #include <stdio.h> #include <string.h> #include <queue> using namespace std; bool ism[5][5]; int a[5][5]; int dx[4]={0, 1, 0, -1}; int dy[4] = { 1, 0, -1, 0 }; struct Node{ int x; int y; int s; short l[30]; }; bool judge(int x, int y){ if (x < 0 || x >= 5 || y < 0 || y >= 5) return true; if (ism[x][y]) return true; if (a[x][y] == 1) return true; return false; } Node bfs(){ queue<Node> q; Node cur, next; cur.x = 0; cur.y = 0; cur.s = 0; ism[cur.x][cur.y] = true; q.push(cur); while (!q.empty()){ cur = q.front(); q.pop(); if (cur.x == 4 && cur.y == 4) return cur; int i, nx, ny; for (i = 0; i < 4; i++){ nx = cur.x + dx[i]; ny = cur.y + dy[i]; if (judge(nx, ny)) continue; next = cur; next.x = nx; next.y = ny; next.s = cur.s+1; next.l[cur.s] = i; q.push(next); } } return cur; } int main(){ int i, j; for (i = 0; i < 5; i++){ for (j = 0; j < 5; j++){ scanf("%d", &a[i][j]); } } memset(ism, 0, sizeof(ism)); Node ans = bfs(); int x, y; x = 0, y = 0; for (i = 0; i<= ans.s; i++){ printf("(%d,%d)\n", x, y); x += dx[ans.l[i]]; y += dy[ans.l[i]]; } return 0; }
标签：要求 inpu tom push mes set eof main ||
原文地址：https://www.cnblogs.com/jianqiao123/p/10792186.html
