！ Gym 100625J The minimum number of open doors for two prisoners-bfs-(priority queue +bfs, the distance from each intersection)

Test instructions: Two-dimensional matrix, the guards from the outside to the inside to put two prisoners, asked the middle of the door to open the minimum number of times.

Analysis:

This question from the outside, so long as the matrix edge can go to the point (except the point of the wall) can be used as a starting point, there are two end points, so the direct lifting point of the search is not feasible. The practice is to use three times BFS, respectively, from the outside to each can walk the minimum distance (number of open doors), two prisoners to each of the minimum distance to go, and then traverse the matrix, the three distance add up, update the answer can be. The minimum distance from the matrix to the matrix is treated like this: add a lap to the input matrix and start the BFS from the (0,0) position (this is also because the intersection of the optimal solution may not be inside the input matrix, but outside).

Because the minimum distance here is not the number of steps to go, but the number of open doors, so BFS to use priority queue, otherwise it will WA.

Code:

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include < string> #include <queue> #define INF 1000000007using namespace Std;int t,n,m;char a[200][200];int d[4][2]={{- 1,0},{1,0},{0,-1},{0,1}};int vis[200][200];int dis[4][200][200];struct node{int x,y;node (int x,int y): x (x), Y (y) {};}; struct Qnode{int x,y;int d;qnode (int x,int y,int D): X (x), Y (y), d (d) {}friend bool operator< (const Qnode &a,const qno De &b) {return a.d>b.d;}}; priority_queue<qnode> q;void bfs (int x,int y,int p) {memset (vis,0,sizeof (VIS)), while (!q.empty ()) Q.pop (); Qnode Tp=qnode (x,y,0); Q.push (TP); Vis[x][y]=1;while (!q.empty ()) {Qnode tmp=q.top (); Q.pop (); for (int i=0;i<4;i++) {int dx =tmp.x+d[i][0];int Dy=tmp.y+d[i][1];if (dx>=0&&dx<=n+1&&dy>=0&&dy<=m+1& &a[dx][dy]!= ' * ' &&!vis[dx][dy]) {vis[dx][dy]=1;if (a[dx][dy]== ' # ') dis[p][dx][dy]=dis[p][tmp.x][tmp.y] +1;else Dis[p][dx][dy]=dis[p][tmp.x][tmp.y];qnode Tmp1=qnode(Dx,dy,dis[p][dx][dy]); Q.push (TMP1);}}} int main () {scanf ("%d", &t), while (t--) {scanf ("%d%d", &n,&m); int ex1=-1,ey1=-1,ex2,ey2;for (int i=0;i<= n+1;i++) a[i][0]= '. ', a[i][m+1]= '. '; for (int i=0;i<=m+1;i++) a[0][i]= '. ', a[n+1][i]= '. for (int i=1;i<=n;i++) scanf ("%s", a[i]+1), and for (int i=1;i<=n;i++) {for (int j=1;j<=m;j++) {if (a[i][j]== ' $ ') {if ( Ex1==-1) Ex1=i,ey1=j;else ex2=i,ey2=j;}}} memset (dis,0,sizeof (DIS)), BFS (0,0,0), BFS (ex1,ey1,1),//for (int i=0;i<n;i++) {//for (int j=0;j<m;j++) cout< <dis[1][i][j];cout<<endl;} BFS (ex2,ey2,2); int ans=inf;for (int i=0;i<=n+1;i++) {for (int j=0;j<=m+1;j++) {if (a[i][j]== '. ' | | a[i][j]== ' $ ') {int sum=0;for (int k=0;k<3;k++) {sum+=dis[k][i][j];} Ans=min (ans,sum);} else if (a[i][j]== ' # ') {int sum=0;for (int k=0;k<3;k++) {sum+=dis[k][i][j];} Ans=min (ans,sum-2);}}} Cout<<ans<<endl;}}

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！ Gym 100625J The minimum number of open doors for two prisoners-bfs-(priority queue +bfs, the distance from each intersection)