目录
[P1746 离开中山路](#P1746 离开中山路)
[P1443 马的遍历](#P1443 马的遍历)
[P1747 好奇怪的游戏](#P1747 好奇怪的游戏)
[P2385 [USACO07FEB] Bronze Lilypad Pond B](#P2385 [USACO07FEB] Bronze Lilypad Pond B)
P1746 离开中山路
cpp
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
int n;
int startx, starty;
int endx, endy;
char a[1010][1010];
int dis[1010][1010];
queue<pair<int, int>> q;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, -1, 0, 1};
int bfs(int x, int y)
{
memset(dis, -1, sizeof dis);
dis[x][y] = 0;
q.push({x, y});
while (!q.empty())
{
auto t = q.front();
q.pop();
for (int i = 0; i < 4; i++)
{
int nx = t.first + dx[i];
int ny = t.second + dy[i];
if (nx >= 1 && ny >= 1 && nx <= n && ny <= n && a[nx][ny] != '1' && dis[nx][ny] <= 0)
{
q.push({nx, ny});
dis[nx][ny] = dis[t.first][t.second] + 1;
}
}
}
return dis[endx][endy];
}
int main()
{
cin >> n;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
cin >> a[i][j];
}
}
cin >> startx >> starty >> endx >> endy;
cout << bfs(startx, starty);
return 0;
}
P1443 马的遍历
2个样例TLE：
cpp
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
int dx[] = {1, 2, 2, 1, -1, -2, -2, -1};
int dy[] = {2, 1, -1, -2, -2, -1, 1, 2};
int dis[410][410];
int n, m;
queue<pair<int, int>> q;
int startx, starty;
int bfs(int x, int y)
{
if (x == startx && y == starty)
{
return 0;
}
memset(dis, -1, sizeof dis);
q.push({startx, starty});
dis[startx][starty] = 0;
while (!q.empty())
{
auto t = q.front();
q.pop();
for (int i = 0; i < 8; i++)
{
int nx = t.first + dx[i];
int ny = t.second + dy[i];
if (nx >= 1 && ny >= 1 && nx <= n && ny <= m && dis[nx][ny] <= 0)
{
q.push({nx, ny});
dis[nx][ny] = dis[t.first][t.second] + 1;
}
}
}
return dis[x][y];
}
int main()
{
cin >> n >> m >> startx >> starty;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
cout << bfs(i, j) << " ";
}
cout << endl;
}
return 0;
}
作者一开始想的是每个点都要计数，所以每个点都要搜一次，然后返回一个值
其实只要搜一次
用void类型，不用返回，从开始搜到结尾，然后每个点都会一层一层地搜到，最后dis数组里存的就是每个点的计数
优化：
cpp
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
int dx[] = {1, 2, 2, 1, -1, -2, -2, -1};
int dy[] = {2, 1, -1, -2, -2, -1, 1, 2};
int dis[410][410];
int n, m;
queue<pair<int, int>> q;
int startx, starty;
void bfs(int x, int y)
{
memset(dis, -1, sizeof dis);
q.push({startx, starty});
dis[startx][starty] = 0;
while (!q.empty())
{
auto t = q.front();
q.pop();
for (int i = 0; i < 8; i++)
{
int nx = t.first + dx[i];
int ny = t.second + dy[i];
if (nx >= 1 && ny >= 1 && nx <= n && ny <= m && dis[nx][ny] <= 0)
{
q.push({nx, ny});
dis[nx][ny] = dis[t.first][t.second] + 1;
}
}
}
}
int main()
{
cin >> n >> m >> startx >> starty;
bfs(startx, starty);
dis[startx][starty] = 0;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
cout << dis[i][j] << " ";
}
cout << endl;
}
return 0;
}
调用STL中的queue调用坐标需要花费比较长的时间，我们可以用数组来模拟queue节省时间
再优化：
cpp
#include <iostream>
#include <cstring>
using namespace std;
int dx[] = {1, 2, 2, 1, -1, -2, -2, -1};
int dy[] = {2, 1, -1, -2, -2, -1, 1, 2};
int dis[410][410];
int n, m;
pair<int, int> q[410 * 410];
int startx, starty;
void bfs(int x, int y)
{
memset(dis, -1, sizeof dis);
int head = 0;
int tail = 0;
q[0] = {x, y};
dis[x][y] = 0;
while (head <= tail)
{
auto t = q[head++];
for (int i = 0; i < 8; i++)
{
int nx = t.first + dx[i];
int ny = t.second + dy[i];
if (nx >= 1 && ny >= 1 && nx <= n && ny <= m && dis[nx][ny] <= 0)
{
q[++tail] = {nx, ny};
dis[nx][ny] = dis[t.first][t.second] + 1;
}
}
}
}
int main()
{
cin >> n >> m >> startx >> starty;
bfs(startx, starty);
dis[startx][starty] = 0;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
cout << dis[i][j] << " ";
}
cout << endl;
}
return 0;
}
好像看不出什么区别
P1747 好奇怪的游戏
cpp
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
int x1, y1;
int x2, y2;
int dis[30][30];
queue<pair<int, int>> q;
int dx[] = {1, 2, 2, 1, -1, -2, -2, -1, 2, 2, -2, -2};
int dy[] = {2, 1, -1, -2, -2, -1, 1, 2, 2, -2, -2, 2};
int bfs(int x, int y)
{
memset(dis, -1, sizeof dis);
q.push({x, y});
dis[x][y] = 0;
while (!q.empty())
{
auto t = q.front();
q.pop();
for (int i = 0; i < 12; i++)
{
int nx = t.first + dx[i];
int ny = t.second + dy[i];
if (nx >= 1 && ny >= 1 && nx <= 20 && ny <= 20 && dis[nx][ny] <= 0)
{
q.push({nx, ny});
dis[nx][ny] = dis[t.first][t.second] + 1;
}
}
}
return dis[1][1];
}
int main()
{
cin >> x1 >> y1 >> x2 >> y2;
cout << bfs(x1, y1) << endl;
cout << bfs(x2, y2) << endl;
return 0;
}
P2385 [USACO07FEB] Bronze Lilypad Pond B
cpp
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
queue<pair<int, int>> q;
int m, n;
int m1, n1;
int a[35][35];
int dis[35][35];
int startx, starty, endx, endy;
int bfs(int x, int y)
{
memset(dis, -1, sizeof dis);
q.push({x, y});
dis[x][y] = 0;
while (!q.empty())
{
auto t = q.front();
q.pop();
int dx[] = {m1, m1, -m1, -m1, n1, n1, -n1, -n1};
int dy[] = {n1, -n1, n1, -n1, m1, -m1, m1, -m1};
for (int i = 0; i < 8; i++)
{
int nx = t.first + dx[i];
int ny = t.second + dy[i];
if (nx >= 1 && ny >= 1 && nx <= m && ny <= n && a[nx][ny] != 0 && a[nx][ny] != 2 && dis[nx][ny] == -1)
{
dis[nx][ny] = dis[t.first][t.second] + 1;
q.push({nx, ny});
}
}
}
return dis[endx][endy];
}
int main()
{
cin >> m >> n >> m1 >> n1;
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
cin >> a[i][j];
if (a[i][j] == 3)
{
startx = i;
starty = j;
}
if (a[i][j] == 4)
{
endx = i;
endy = j;
}
}
}
cout << bfs(startx, starty);
return 0;
}
