我使用了递归的dfs没有通过,“我的提交里”看到测试用例只有一个,结果是输出超时,估计是递归法使得复杂度陡增,这是题目:
虽然0分,我还是保存一下代码供以后参考:
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
bool flag;
bool testunit[30][30];
int n, m;
char map[30][30];
int ex, ey;
int sx, sy;
int firex, firey;
int minn;
int dr[4] = { -1, 0, 1, 0 };
int dc[4] = { 0, 1, 0, - 1 };
float distance(int x1, int y1, int x2, int y2)
{
return sqrt(pow(x1 - x2, 2) + pow(y1 - y2, 2));
}
void dfs(int x,int y,int cnt,int flag1,int flag2)
{
if (x == ex && y == ey)
{
flag = true;
if (cnt < minn)minn = cnt;
return;
}
for (int i = 0; i < 4; i++)
{
if (i == flag1 || i == flag2)continue;
if (x + dr[i] >= 0 && x + dr[i] < n&&y + dc[i] >= 0 && y + dc[i] < m&&map[x + dr[i]][y + dc[i]] != '#'&&!testunit[x + dr[i]][y + dc[i]])
{
testunit[x + dr[i]][y + dc[i]] = true;
dfs(x + dr[i], y + dc[i], cnt + 1, flag1, flag2);
testunit[x + dr[i]][y + dc[i]] = false;
}
}
return;
}
int main(void)
{
memset(testunit, false, sizeof(testunit));
int N;
cin >> N;
for (int k = 0; k < N; k++)
{
flag = false;
minn = 100000;
cin >> n >> m;
for (int i = 1; i <= n; i++)
for (int j = 0; j < m; j++)
{
cin >> map[i][j];
if (map[i][j] == 'E') { ex = i; ey = j; }
if (map[i][j] == 'S') { sx = i; sy = j; }
if (map[i][j] == '*') { firex = i; firey = j; }
}
if (distance(ex, ey, sx, sy) < (firex, firey, ex, ey))
{
int flag1, flag2;
if (ex >= sx)flag1 = 3;
else flag1 = 1;
if (ey >= sy)flag2 = 0;
else flag2 = 2;
dfs(sx, sy, 0,flag1,flag2);
}
testunit[sx][sy] = true;
if (flag == false)cout << "T_T" << endl;
else cout << minn << endl;
}
system("pause");
return 0;
}
于是又试了BFS,代码如下:
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
struct node
{
int x, y;
int step;
}S,E,fire,Node;
bool flag;
bool testunit[35][35];
int n, m;
char map[35][35];
int dr[4] = { -1, 0, 1, 0 };
int dc[4] = { 0, 1, 0, -1 };
float distance(int x1, int y1, int x2, int y2)
{
return sqrt(pow(x1 - x2, 2) + pow(y1 - y2, 2));
}
bool test(int x, int y)
{
if (x < 0 && x >= n && y < 0 && y >= m)return false;
if (testunit[x][y] == true)return false;
if (map[x][y] == '#')return false;
return true;
}
int bfs()
{
queue<node> Q;
Q.push(S);
while (!Q.empty())
{
node top = Q.front();
Q.pop();
if (top.x == E.x && top.y == E.y)return top.step;
for (int i = 0; i < 4; i++)
{
int X = top.x + dr[i];
int Y = top.y + dc[i];
if (test(X, Y))
{
Node.x = X;
Node.y = Y;
Node.step = top.step + 1;
Q.push(Node);
testunit[X][Y] = true;
}
}
}
return -1;
}
int main(void)
{
memset(testunit, false, sizeof(testunit));
memset(map, '#', sizeof(map));
int N;
cin >> N;
for (int k = 0; k < N; k++)
{
flag = false;
cin >> n >> m;
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
{
char temp;
cin >> temp;
map[i][j] = temp;
if (map[i][j] == 'E') { E.x = i; E.y = j; }
if (map[i][j] == 'S') { S.x = i; S.y = j; }
if (map[i][j] == '*') { fire.x = i; fire.y = j; }
}
int result = -1;
if (distance(E.x, E.y, S.x, S.y) < distance(fire.x, fire.y, E.x, E.y))
{
result = bfs();
}
testunit[S.x][S.y] = true;
if (result == -1)cout << "T_T" << endl;
else cout << result << endl;
}
system("pause");
return 0;
}
结果显示段错误,应该是循环多次导致栈溢出,我于是看了一个AC的代码,他用了回溯法,我把BFS那边的注释写了下,如下所示,是可以AC的:
#include<iostream>
#include<cmath>
#include<queue>
using namespace std;
int n, m;
int dir[4][2] = { {1,0},{-1,0},{0,1},{0,-1} };
char mp[31][31];
int vis[31][31], f[31][31];
struct node
{
int x;
int y;
int step;
};
int sx, sy, fx, fy;
queue<node>q;
void bfs(int x, int y)
{
queue<node>q;
node a, b, temp1, temp2;
a.x = x;
a.y = y;
a.step = 0;
q.push(a);
while (!q.empty())
{
a = q.front();
q.pop();
//满足条件时
if (mp[a.x][a.y] == 'E')
{
cout << a.step << endl;
return;
}
//这一步是用来简化复杂度的,是回溯算法,
//满足条件的话说明当前循环以前已经运行过了,就跳过当前这一循环。
if (f[a.x][a.y] == a.step) continue;
for (int i = 0; i < 4; i++)
{
//初始化b,其意义不是把a拷贝给b
b = a;
//遍历下一格
b.x = a.x + dir[i][0];
b.y = a.y + dir[i][1];
//这一步好像有点累赘,但是可以看出是用来下面判断是否满足遍历的条件的
int xx = b.x;
int yy = b.y;
//当前加一步
b.step += 1;
//遍历下一格子
if (mp[xx][yy] != '#' && b.step <= f[xx][yy] && !vis[xx][yy] && xx > 0 && xx <= n && yy > 0 && yy <= m)
{
q.push(b);
vis[xx][yy] = 1;
}
}
}
//不满足条件时
cout << "T_T" << endl;
return;
}
int main()
{
int t;
cin >> t;
while (t)
{
while (!q.empty()) q.pop();
memset(vis, 0, sizeof(vis));
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
cin >> mp[i][j];
if (mp[i][j] == 'S')
{
sx = i;
sy = j;
}
if (mp[i][j] == '*')
{
fx = i;
fy = j;
}
}
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
{
f[i][j] = max(abs(fx - i), abs(fy - j));
}
vis[sx][sy] = 1;
bfs(sx, sy);
t = t - 1;
}
}