HDU 4856 tunnels (BFS + pressure DP)

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Question:

A square grid graph with a side length of N, with some vertices '. 'indicates reachable,' # 'indicates inaccessible. You cannot go to an inaccessible point, and you can only go to an adjacent grid (up, down, and left) per unit of time ). Now we will give you m channels, starting positions (x1, Y1) and ending points (X2, Y2) of each channel ), when you get in from the starting point, you can immediately get out of the ending point (no time is spent), but you can only go through each secret path. Now, you can select any reachable point as the starting point and ask if you can finish all the secrets if the conditions are met. If there is a solution to output the shortest time, otherwise-1 is output.

Analysis:

The distance between each tunnel is first processed by BFs, and then the order between the tunnels is followed. The pressure DP can be used for processing,

DP [I] [J] indicates that the passed channel status is I (I indicates the compression status, and each bit of binary indicates whether the corresponding channel has passed, and the pass is 1, otherwise, it is 0), and the last pass is the minimum time when J is used.

The state transition equation is: DP [I | (1 <k)] [k] = min {DP [I | (1 <k)] [K], DP [I] [J] + dist [J] [k]}.
Ensure that the DP [I] [J] is in a state. Dist [J] [k] is not Inf (J can reach K ), and status I does not include the K th secret channel before transfer.

 1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <queue> 5 #include <cstring> 6 #include <cstdlib> 7 #include <algorithm> 8 #define LL __int64 9 const int maxn = 20+10;10 using namespace std;11 const int INF = 1<<28;12 char s[maxn][maxn];13 int c[maxn][maxn], d[1<<17][20], n;14 int dx[] = {0, 0, 1, -1};15 int dy[] = {1, -1, 0, 0};16 struct node17 {18     int x, y, step;19 }pos, ne;20 int bfs(int a, int b, int c, int d)21 {22     queue<node>q;23     int vis[maxn][maxn], i;24     memset(vis, 0, sizeof(vis));25     ne.x = a; ne.y = b; ne.step = 0;26     vis[a][b] = 1;27     q.push(ne);28     while(!q.empty())29     {30         pos = q.front();31         q.pop();32         if(pos.x == c && pos.y == d)33         return pos.step;34         for(i = 0; i < 4; i++)35         {36             ne.x = pos.x+dx[i];37             ne.y = pos.y+dy[i];38             ne.step = pos.step+1;39             if(!(ne.x>=1&&ne.x<=n && ne.y>=1&&ne.y<=n)) continue;40             if(s[ne.x][ne.y]==‘#‘) continue;41             if(vis[ne.x][ne.y]) continue;42             q.push(ne);43             vis[ne.x][ne.y] = 1;44         }45     }46     return INF;47 }48 int main()49 {50     int m, i, j, k, ans;51     int x1[maxn], y1[maxn], x2[maxn], y2[maxn];52     while(~scanf("%d%d", &n, &m))53     {54         memset(c, 0, sizeof(c));55         memset(s, 0, sizeof(s));56         for(i = 1; i <= n; i++)57         {58             getchar();59             for(j = 1; j <= n; j++)60             scanf("%c", &s[i][j]);61         }62         for(i = 0; i < m; i++)63          cin>>x1[i]>>y1[i]>>x2[i]>>y2[i];64         for(i = 0; i < m; i++)65         for(j = 0; j < m; j++)66         {67             if(i == j) continue;68             c[i][j] = bfs(x2[i], y2[i], x1[j], y1[j]);69         }70         for(i = 0; i < (1<<m); i++)71         for(j = 0; j < m; j++)72         d[i][j] = INF;73 74         for(i = 0; i < m; i++)75         d[1<<i][i] = 0;76 77         for(i = 0; i < (1<<m); i++)78         for(j = 0; j < m; j++)79         if(d[i][j]!=INF)80         for(k = 0; k < m; k++)81         {82             if(!(i&(1<<k)) && c[j][k]!=INF)83             d[i|(1<<k)][k] = min(d[i|(1<<k)][k], d[i][j]+c[j][k]);84         }85         ans = INF;86         for(i = 0; i < m; i++)87         ans = min(ans, d[(1<<m)-1][i]);88         if(ans == INF) printf("-1\n");89         else90         printf("%d\n", ans);91     }92     return 0;93 }

 

HDU 4856 tunnels (BFS + pressure DP)