http://poj.org/problem?id=2251
题意:给你一个三维的地图,找一条最短路
题解:因为是最短路,所以宽搜。三维就用六个单位向量。
坑:
第一次打bfs各种打错:
首先是字符数组处理: 一行一行读可以用cin,但这样就不能map[i][j]+1读入。(for i=1 or i=0;)
数组初始定义,int定义成char了,memset也错了
忘记pop,
return now not v;
l,r,c,x,y,z,对应关系弄混
测试的时候改了一行忘记改回来,
于是一直卡样例Orz
ac代码:
#define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<vector> #include<algorithm> #include<stdio.h> #include<queue> #include<string.h> using namespace std; const int maxn = 1e5 + 5; char map[35][35][35]; int step[35][35][35];// bool vis[35][35][35]; int l, r, c; int dir[6][3] = { 1,0,0, -1,0,0, 0,1,0, 0,-1,0, 0,0,1, 0,0,-1 }; struct point { int x, y, z; point(int x=0, int y=0, int z=0) :x(x), y(y),z(z) {} }s,now,v; //bool check(point q) { // if (!vis[q.z][q.x][q.y] && q.z >= 0 && q.z < l&&q.x >= 0 && q.x < r&&q.y >= 0 && q.y < c&&map[q.z][q.x][q.y] != ‘#‘) // return 1; // return 0; //} int check(int x, int y, int z) { if (!vis[z][x][y] && z >= 0 && z < l&&x >= 0 && x < r&&y >= 0 && y < c&&map[z][x][y] != ‘#‘)// return 1; return 0; } int bfs() { memset(vis, 0, sizeof(vis)); memset(step, 0, sizeof(step));// vis[s.z][s.x][s.y] = 1; step[s.z][s.x][s.y] = 0; //step[s.] queue<point> Q; Q.push(s); while (!Q.empty()) { now = Q.front(); Q.pop();// for (int i = 0; i < 6; i++) { int dz = now.z + dir[i][0]; int dx = now.x + dir [i][1]; int dy = now.y + dir[i][2]; v = point( dx, dy,dz);// if (check(v.x,v.y,v.z)) { if (map[v.z][v.x][v.y] == ‘E‘)return step[now.z][now.x][now.y] + 1;// Q.push(v); vis[v.z][v.x][v.y] = 1; step[v.z][v.x][v.y] = step[now.z][now.x][now.y] + 1; } } } return -1; }; int main() { while (cin >> l >> r >> c) { if (!l && !r && !c) break; for (int i = 0; i <l; i++) for (int j = 0; j < r; j++)scanf("%s", map[i][j]);//cin >> map[i][j]; for (int i = 0; i < l; i++) for (int j =0; j < r; j++) for (int k = 0; k < c; k++) { if (map[i][j][k] == ‘S‘) {// s=point(j,k,i); } } //cout << s.y << endl; int ans = bfs(); ans == -1 ? printf("Trapped!\n"): printf("Escaped in %d minute(s).\n", ans); } }