NYOJ58 最少步数 【深搜】+【广搜】

最少步数

时间限制：3000 ms  |  内存限制：65535 KB

难度：4

描述

这有一个迷宫，有0~8行和0~8列：

 1,1,1,1,1,1,1,1,1
 1,0,0,1,0,0,1,0,1
 1,0,0,1,1,0,0,0,1
 1,0,1,0,1,1,0,1,1
 1,0,0,0,0,1,0,0,1
 1,1,0,1,0,1,0,0,1
 1,1,0,1,0,1,0,0,1
 1,1,0,1,0,0,0,0,1
 1,1,1,1,1,1,1,1,1

0表示道路，1表示墙。

现在输入一个道路的坐标作为起点，再如输入一个道路的坐标作为终点，问最少走几步才能从起点到达终点？

（注：一步是指从一坐标点走到其上下左右相邻坐标点，如：从（3，1）到（4,1）。）

输入

第一行输入一个整数n（0<n<=100），表示有n组测试数据;
随后n行,每行有四个整数a,b,c,d（0<=a,b,c,d<=8）分别表示起点的行、列，终点的行、列。

输出

输出最少走几步。

样例输入

23 1  5 73 1  6 7

样例输出

1211

广搜：

#include <stdio.h>#include <string.h>#include <queue>using std::queue;bool map[9][9] = { 1,1,1,1,1,1,1,1,1, 1,0,0,1,0,0,1,0,1, 1,0,0,1,1,0,0,0,1, 1,0,1,0,1,1,0,1,1, 1,0,0,0,0,1,0,0,1, 1,1,0,1,0,1,0,0,1, 1,1,0,1,0,1,0,0,1, 1,1,0,1,0,0,0,0,1, 1,1,1,1,1,1,1,1,1 }, vis[9][9];  int ax, ay, bx, by; int mov[][2] = {0, 1, 0, -1, -1, 0, 1, 0}; struct Node{int x, y, steps; };  queue<Node> Q;  void BFS(){Node temp, now;while(!Q.empty()){now = Q.front();Q.pop();if(now.x == bx && now.y == by){printf("%d\n", now.steps);return;}++now.steps;for(int i = 0; i < 4; ++i){temp = now;temp.x += mov[i][0];temp.y += mov[i][1];if(!vis[temp.x][temp.y] && !map[temp.x][temp.y]){vis[temp.x][temp.y] = 1;Q.push(temp);}}} }  int main(){int n;scanf("%d", &n);while(n--){memset(vis, 0, sizeof(vis));scanf("%d%d%d%d", &ax, &ay, &bx, &by);Node temp; temp.x = ax;temp.y = ay; temp.steps = 0;while(!Q.empty()) Q.pop();Q.push(temp);BFS();}return 0; }

深搜：

 #include <stdio.h>bool sam[9][9] = { 1,1,1,1,1,1,1,1,1, 1,0,0,1,0,0,1,0,1, 1,0,0,1,1,0,0,0,1, 1,0,1,0,1,1,0,1,1, 1,0,0,0,0,1,0,0,1, 1,1,0,1,0,1,0,0,1, 1,1,0,1,0,1,0,0,1, 1,1,0,1,0,0,0,0,1, 1,1,1,1,1,1,1,1,1 };  int a, b, c, d, minlen, len; int f[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};  void DFS(int x, int y){if(len >= minlen) return;if(x == c && y == d){ minlen = len; return; }for(int i = 0; i < 4; ++i){if(sam[x+f[i][0]][y+f[i][1]] == 0){sam[x+f[i][0]][y+f[i][1]] = 1;++len;DFS(x+f[i][0], y+f[i][1]);--len;sam[x+f[i][0]][y+f[i][1]] = 0;}} }  int main(){int t;scanf("%d", &t);while(t--){scanf("%d%d%d%d", &a, &b, &c, &d);minlen = 100;len = 0;DFS(a, b);printf("%d\n", minlen);}return 0; }