HDU 4856 (bfs 状压DP)

Tunnels

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1925    Accepted Submission(s): 570

Problem Description

Bob is travelling in Xi’an. He finds many secret tunnels beneath the city. In his eyes, the city is a grid. He can’t enter a grid with a barrier. In one minute, he can move into an adjacent grid with no barrier. Bob is full of curiosity and he wants to visit all of the secret tunnels beneath the city. To travel in a tunnel, he has to walk to the entrance of the tunnel and go out from the exit after a fabulous visit. He can choose where he starts and he will travel each of the tunnels once and only once. Now he wants to know, how long it will take him to visit all the tunnels (excluding the time when he is in the tunnels).

 

Input

The input contains mutiple testcases. Please process till EOF.
For each testcase, the first line contains two integers N (1 ≤ N ≤ 15), the side length of the square map and M (1 ≤ M ≤ 15), the number of tunnels.
The map of the city is given in the next N lines. Each line contains exactly N characters. Barrier is represented by “#” and empty grid is represented by “.”.
Then M lines follow. Each line consists of four integers x ₁, y ₁, x ₂, y ₂, indicating there is a tunnel with entrence in (x ₁, y ₁) and exit in (x ₂, y ₂). It’s guaranteed that (x ₁, y ₁) and (x ₂, y ₂) in the map are both empty grid.

 

Output

For each case, output a integer indicating the minimal time Bob will use in total to walk between tunnels.
If it is impossible for Bob to visit all the tunnels, output -1.

 

Sample Input

   
   
   
   

    
    
    
    5 4
....#
...#.
.....
.....
.....
2 3 1 4
1 2 3 5
2 3 3 1
5 4 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    7
   
   
   
   

 

题意:有一个图,#表示这个格子无法走到,然后格子之间移动一格花费一分钟,通道之间移动不花费时间,求通

过所有的通道花费的最小时间.

首先bfs预处理每个通道终点到所有点的最短距离,然后用dp[i][j]表示通道经过状态为i下在j通道终点的最少

花费,那么dp[i][j]=min(dp[i^(1<<j)][k]+dis (k,j)).

坑点是不能通过所有通道的情况的判断.

#include <bits/stdc++.h>
using namespace std;
#define maxn 16
#define maxm 1<<16
#define INF 11111111

int dis[333][333];
int n, m, Max;
char mp[maxn][maxn];
int s[maxn][2], t[maxn][2];//起点终点
int dp[maxm][maxn];

int id (int x, int y) {
    return x*n+y;
}

bool vis[maxn][maxn];
#define move Move
const int move[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
bool legal (int x, int y) {
    if (x < 0 || y < 0 || x >= n || y >= n)
        return 0;
    if (mp[x][y] == '#')
        return 0;
    return 1;
}

void bfs (int x, int y) { 
    queue <int> gg;
    while (!gg.empty ())
        gg.pop();
    memset (vis, 0, sizeof vis);
    gg.push (id (x, y));
    vis[x][y] = 1;
    dis[id(x,y)][id(x,y)] = 0;
    while (!gg.empty ()) {
        int now = gg.front (); gg.pop ();
        int nowx = now/n, nowy = now%n;
        for (int i = 0; i < 4; i++) {
            int xx = nowx+move[i][0], yy = nowy+move[i][1];
            if (!legal (xx, yy) || vis[xx][yy])
                continue;
            dis[id(x, y)][id (xx, yy)] = dis[id(x, y)][id (nowx, nowy)]+1;
            vis[xx][yy] = 1;
            gg.push (id (xx, yy));
        }
    }
    return ;
}

vector <int> num[maxn];
#define count Count
int count (int num) {
    int ans = 0;
    while (num) {
        if (num&1)
            ans++;
        num >>= 1;
    }
    return ans;
}
void init () {
    for (int i = 0; i <= m; i++)
        num[i].clear ();
    for (int i = 0; i < Max; i++) {
        num[count (i)].push_back (i);
    }
    /*for (int i = 0; i <= m; i++) {
        for (int j = 0; j < num[i].size (); j++) {
            cout << num[i][j] << " ";
        }  cout << endl;
    }*/
    return ;
}

int main () {
    //freopen ("in", "r", stdin);
    while (scanf ("%d%d", &n, &m) == 2) {
        memset (dis, -1, sizeof dis);
        for (int i = 0; i < n; i++) {
            scanf ("%s", mp[i]);
        }
        Max = (1<<m);
        for (int i = 0; i < m; i++) {
            scanf ("%d%d%d%d", &s[i][0], &s[i][1], &t[i][0], &t[i][1]);
            s[i][0]--, s[i][1]--, t[i][0]--, t[i][1]--;
            bfs (t[i][0], t[i][1]);
        }
        for (int i = 0; i < Max; i++) {
            for (int j = 0; j < m; j++)
                dp[i][j] = INF;
        }
        for (int i = 1, j = 0; i < Max; i <<= 1, j++) {
            dp[i][j] = 0;
        }
        init ();
        for (int cnt = 2; cnt <= m; cnt++) {
            for (int l = 0; l < num[cnt].size (); l++) {
                int i = num[cnt][l]; 
                for (int j = 0; j < m; j++) if (i&(1<<j)) {
                    for (int k = 0; k < m; k++) if ((1<<k)&(i^(1<<j))) { 
                        if (dis[id(t[k][0],t[k][1])][id(s[j][0],s[j][1])] == -1)
                            continue;
                        dp[i][j] = min (dp[i][j], dp[i^(1<<j)][k]+dis[id(t[k][0],t[k][1])][id(s[j][0],s[j][1])]);
                    }
                }
            }
        } 
        int ans = INF;
        for (int i = 0; i < m; i++) {
            ans = min (ans, dp[Max-1][i]);
        }
        if (ans >= INF)
            printf ("-1\n");
        else
            printf ("%d\n", ans);
    }
    return 0;
}