说明：
基本都是bfs的常见模板题型，思路都很直接，不过后面有两道题很搞心态，它们给的坐标x、y是反的，导致刚开始一直错。题目还是要看仔细，不能先入为主。
题目列表：
问题 A: 围圈报数(完善程序)
参考题解：
#include<iostream>
#include<queue>
using namespace std;
int n,m,k=1,tmp;
queue<int> arr;
int main()
{cin>>n>>m;for(int i=1;i<=n;i++)arr.push(i);// _____(1)____//依次进入队列while(arr.size())// while(_____(2)_____)//判断队列里是否还有人{tmp=arr.front();if(k%m==0)cout<<tmp<<" ";elsearr.push(tmp);// ______(3)______//如果不是第m个人，则重新入队// _____(4)_____//从队列里删除arr.pop();k++;}return 0;
}问题 B: 围圈报数
参考题解：
#include <iostream>
#include <queue>
using std::cin;
using std::cout;
int main() {cin.tie(nullptr)->sync_with_stdio(false);int n,m,count = 0;cin >> n >> m;std::queue<int> q;for(int i = 1;i<=n;i++) q.push(i);while(q.size()){auto t = q.front();count++;if(count%m==0) {cout << t << ' ';}else {q.push(t);}q.pop();}cout << std::endl;return 0;
}问题 C: 报数相同次数circle
参考题解：
#include <iostream>
#include <queue>
using std::cin;
using std::cout;
int main() {cin.tie(nullptr)->sync_with_stdio(false);int n;cin >> n;int a,b;cin >> a >> b;std::queue<int> q1,q2;for(int i = 1;i<=a;i++) q1.push(i);for(int i = 1;i<=b;i++) q2.push(i);int count = 0;for(int i = 1;i<=n;i++) {auto t1 = q1.front(),t2 = q2.front();if(t1==t2) count++;q1.push(t1),q2.push(t2);q1.pop(),q2.pop();}cout << count << std::endl;return 0;
}问题 D: 最小倍数
参考题解：
#include <iostream>
#include <queue>
using std::cin;
using std::cout;
using ll = long long;
ll n,x;
void bfs() {std::queue<ll> q;q.push(1);while(q.size()) {x = q.front();q.pop();if(x%n==0&&x>=n) {cout << x << '\n';return;}q.push(x*10);q.push(x*10+1);}cout << x << '\n';
}
int main() {cin.tie(nullptr)->sync_with_stdio(false);while(cin >> n,n) {bfs();}return 0;
}问题 E: 迷宫的最短路径
参考题解：
#include <iostream>
#include <queue>
#include <cstring>
using std::cin;
using std::cout;
int main() {cin.tie(nullptr)->sync_with_stdio(false);constexpr int N = 1e2+5;int sx = 1,sy = 1,ex = 1,ey = 1;struct node {int x,y,s;}t,t1;char g[N][N];bool vis[N][N];memset(vis,false,sizeof vis);int dx[] = {0,0,-1,1};int dy[] = {-1,1,0,0};std::queue<node> q;int n,m;cin >> n >> m;for(int i = 1;i<=n;i++) {for(int j = 1;j<=m;j++) {cin >> g[i][j];if(g[i][j]=='S') {sx=i,sy=j;}else if(g[i][j]=='G') {ex=i,ey=j;}}}auto bfs = [&]()->void{t.x = sx,t.y = sy,t.s = 0;q.push(t);vis[sx][sy] = true;while(!q.empty()) {t = q.front();q.pop();if(t.x==ex&&t.y==ey) {cout << "The min steps are:" << t.s << "!\n";return;}for(int i = 0;i<4;i++) {int u = t.x+dx[i],v = t.y+dy[i];if(u<1||u>n||v<1||v>m||vis[u][v]||g[u][v]=='#') continue;vis[u][v] = true;t1.x = u,t1.y = v,t1.s = t.s+1;q.push(t1);}}cout << "sorry!\n";};bfs();return 0;
}问题 F: 象棋中的马之进阶
参考题解：
#include <iostream>
#include <queue>
#include <cstring>
using std::cin;
using std::cout;
int main() {cin.tie(nullptr)->sync_with_stdio(false);constexpr int N = 15;struct node {int x,y,s;}t,t1;int dx[] = {-1,1,2,2,1,-1,-2,-2};int dy[] = {2,2,1,-1,-2,-2,-1,1};int sx,sy,ex,ey;bool vis[N][N];memset(vis,false,sizeof vis);cin >> sy >> sx >> ey >> ex;std::queue<node> q;auto bfs = [&]() ->void {t.x = sx,t.y = sy,t.s = 0;vis[sx][sy] = true;q.push(t);while(!q.empty()) {t = q.front();q.pop();if(t.x==ex&&t.y==ey) {cout << t.s << std::endl;return;}for(int i = 0;i<8;i++) {int u = t.x+dx[i],v = t.y+dy[i];if(u<1||u>10||v<1||v>9||vis[u][v]) continue;vis[u][v] = true;t1.x = u,t1.y = v,t1.s = t.s+1;q.push(t1);}}cout << 0 << std::endl;};bfs();return 0;
}
问题 G: 迷宫探险
参考题解：
#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
using std::cin;
using std::cout;
int main() {cin.tie(nullptr)->sync_with_stdio(false);constexpr int N = 1e2+5;struct node {int x,y,s;bool operator > (const node &W) const {return s > W.s;}}t,t1;char g[N][N];bool vis[N][N];int dx[] = {0,0,-1,1};int dy[] = {-1,1,0,0};int n,sx = 1,sy = 1,ex = 1,ey = 1;std::priority_queue<node,std::vector<node>,std::greater<node>> pq;auto bfs = [&]() ->void {while(!pq.empty()) pq.pop();memset(vis,false,sizeof vis);t.x = sx,t.y = sy,t.s = 0;vis[sx][sy] = true;pq.push(t);while(!pq.empty()) {t = pq.top();pq.pop();if(t.x==ex&&t.y==ey) {cout << t.s << '\n';return;}for(int i = 0;i<4;i++) {int u = t.x+dx[i],v = t.y+dy[i];if(u<1||u>n||v<1||v>n||vis[u][v]||g[u][v]=='#') continue;vis[u][v] = true;int ds = 1;if(g[u][v]!='.') ds += int(g[u][v]^48);t1.x = u,t1.y = v,t1.s = t.s+ds;pq.push(t1);}}cout << -1 << '\n';};while(cin >> n) {ex = n,ey = n;for(int i = 1;i<=n;i++) {for(int j = 1;j<=n;j++) {cin >> g[i][j];}}bfs();}return 0;
}问题 H: 迷宫
参考题解：
#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
using std::cin;
using std::cout;
int main() {cin.tie(nullptr)->sync_with_stdio(false);constexpr int N = 1e2+5;struct node {int x,y,s;}t,t1;char g[N][N];bool vis[N][N];int dx[] = {0,0,-1,1};int dy[] = {-1,1,0,0};int sx,sy,ex,ey,k,n,m;std::queue<node> q;auto bfs = [&]()->void {memset(vis,false,sizeof vis);while(!q.empty()) q.pop();t.x = sx,t.y = sy,t.s = -1;vis[sx][sy] = true;q.push(t);while(!q.empty()) {t = q.front();q.pop();if(t.x==ex&&t.y==ey&&t.s<=k) {cout << "yes\n";return;}for(int i = 0;i<4;i++) {t1.x = t.x+dx[i],t1.y = t.y+dy[i];int &u = t1.x,&v = t1.y;while(u>=1&&u<=n&&v>=1&&v<=m&&g[u][v]=='.') {if(!vis[u][v]) {t1.s=t.s+1;vis[u][v] = true;q.push(t1);}u+=dx[i],v+=dy[i];}}}cout << "no\n";};int T = 1;cin >> T;while(T--) {cin >> n >> m;for(int i = 1;i<=n;i++) {for(int j = 1;j<=m;j++) {cin >> g[i][j];}}cin >> k >> sy >> sx >> ey >> ex;bfs();}return 0;
}
)
