BFS之最短路径模型
当一个图的每个边的权重都一样的时候,会有一个最短路径模型。不需要考虑边的影响。
1076. 迷宫问题 - AcWing题库
#include<iostream>
#include<queue>
#include<utility>
#include<algorithm>
#include<stack>
using namespace std;
const int N = 1010;
typedef pair<int,int> PII;
int g[N][N]; //存储地图
bool st[N][N]; //存储状态
PII pre[N][N];//存储前一个点
int n;
PII track[N];
int k=0;
int dx[] = {-1,0,1,0};
int dy[] = {0,1,0,-1};void bfs(int sx,int sy){queue<PII> q;q.push(make_pair(sx,sy));st[sx][sy] = true;while(!q.empty()){PII 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 >=0 && nx<n && ny>=0 && ny<n && g[nx][ny]==0 && st[nx][ny]==false){pre[nx][ny] = t;q.push(make_pair(nx,ny));st[nx][ny] = true;}// if(nx<0 || nx>n-1 || ny<0 || ny>n-1) continue;// if(st[nx][ny]) continue;// if(g[nx][ny]==1) continue;// pre[nx][ny] = t;// q.push(make_pair(nx,ny));// st[nx][ny] = true;}}
}int main(){cin>>n;for(int i=0;i<n;i++){for(int j=0;j<n;j++){cin>>g[i][j];pre[i][j] = {-1, -1};}}bfs(0,0);stack<PII> s;PII present = pre[n-1][n-1];while(true){s.push(present);if(present.first==0 && present.second==0) break;present = pre[present.first][present.second];}while(!s.empty()){PII v = s.top();s.pop();cout<<v.first<<" "<<v.second<<endl;}cout<<n-1<<" "<<n-1<<endl;return 0;
}
马走日模型
188. 武士风度的牛 - AcWing题库
#include<iostream>
#include<utility>
#include<queue>
#include<algorithm>
using namespace std;
typedef pair<int,int> PII;
PII start,end;
const int N = 160;
int n,m,res=-1;
char g[N][N];
bool st[N][N];
bool flag;
int dist[N][N];
int dx[] = {-2,-1,1,2,2,1,-1,-2};
int dy[] = {1,2,2,1,-1,-2,-2,-1};void bfs(int sx,int sy){queue<PII> q;q.push(make_pair(sx,sy));st[sx][sy] = true;while(!q.empty()){if(flag) break;PII 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 || nx >n || ny <1 || ny>m) continue;//不能越界if(st[nx][ny]) continue;//遍历过的不能再次遍历if(g[nx][ny]=='*') continue;//不能在*上面if(g[nx][ny]=='H'){ //走到有草的地方了,立即暂停并打印cout<<dist[t.first][t.second]+1;flag = true;break;}q.push(make_pair(nx,ny));st[nx][ny] = true;dist[nx][ny] = dist[t.first][t.second]+1;}}
}int main(){cin>>m>>n;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>g[i][j];if(g[i][j]=='K'){start.first = i;start.second = j;}}}// for(int i=1;i<=n;i++){ //检查输入// for(int j=1;j<=m;j++){// cout<<g[i][j];// }// cout<<endl;// }bfs(start.first,start.second);return 0;
}