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
1,1,1,1,1,1,1,1,10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
-
2 3 1 5 7 3 1 6 7
- 样例输出
-
12 11
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;
}