POJ 3984迷宫问题(BFS)

简单深搜

#include<iostream>#include<cstdio>#include<algorithm>#include<cstdlib>#include<cmath>#include<string.h>#include<cstring>#include<string>#include<map>#include<set>#include<vector>#include<ctime>#include<queue>#define hash hashhusing namespace std;typedef long long ll;#define sp  system("pause")#define PB push_backint G[5][5];int vis[5][5];int dx[4] = { 0, 0, 1, -1 };int dy[4] = { 1, -1, 0, 0 };class node{public:int now;int x, y;int px[26];int py[26];};int main(){memset(vis, 0, sizeof vis);memset(G, 0, sizeof G);for (int i = 0; i < 5; i++){for (int j = 0; j < 5; j++)scanf("%d", &G[i][j]);}node start;start.now = 0;start.x = 0;start.y = 0;start.px[start.now] = 0;start.py[start.now] = 0;queue<node>q;q.push(start);vis[0][0] = 1;while (!q.empty()){node t = q.front();if (t.x == 4 && t.y == 4){for (int i = 0; i <= t.now; i++){printf("(%d, %d)\n", t.px[i], t.py[i]);}sp;return 0;}q.pop();for (int i = 0; i < 4; i++){int a = t.x + dx[i];int b = t.y + dy[i];if (a >= 0 && a< 5 && b>= 0 &&b < 5&&!vis[a][b]&&!G[a][b]){vis[a][b] = 1;node in;in.now = t.now + 1;in.x = t.x + dx[i];in.y = dy[i] + t.y;memcpy(in.px, t.px, sizeof t.px); memcpy(in.py, t.py, sizeof t.py);in.px[in.now] = in.x;in.py[in.now] = in.y;q.push(in);}}}return 0;}