HDU 4856 Tunnels(bfs+状压dp)

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本文介绍了一种利用广度优先搜索(BFS)和动态规划(DP)解决迷宫寻径问题的方法，旨在寻找通过所有秘密隧道所需的最短时间。通过预计算不同隧道间的最短路径并使用状态压缩技巧，实现了高效的全局最优解。

题目大意:给你一个N*N的图让你到达所有的“.”点，“#”不能通过，有m组每组有一个入口，一个出口，入口可以传送到出口，不知道经过m组的先后顺序，让你求出走过所有的“.”的最小时间。

思路：先bfs出来所有的m之间的最短距离，然后dp[j][i] 表示，在j状态下开始第i步的最小路程，枚举找到一个最小的dp[1<<m - 1][i]，就是最短距离，否则输出“-1”。

Tunnels

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1139 Accepted Submission(s): 344

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

#include <set>
#include <map>
#include <queue>
#include <math.h>
#include <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

#define LL __int64


using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 20;

struct node
{
    int x0, y0;
    int x1, y1;
} f[maxn];

struct node1
{
    int x, y;
};

int cnt;

char str[maxn][maxn];
bool vis[maxn][maxn];
int dis[maxn][maxn];
int d[maxn][maxn];
int dx[] = {0, 1, -1, 0};
int dy[] = {1, 0, 0, -1};

int dp[1<<16][maxn];
int mp[maxn][maxn];
int n, m;

void spfa(int s, int t)
{
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            vis[i][j] = false;
            d[i][j] = INF;
        }
    }
    node1 tmp;
    tmp.x = s;
    tmp.y = t;
    vis[s][t] = true;
    queue<node1>fp;
    fp.push(tmp);
    d[s][t] = 0;
    while(!fp.empty())
    {
        tmp = fp.front();
        fp.pop();
        for(int i = 0; i < 4; i++)
        {
            int x = tmp.x+dx[i];
            int y = tmp.y+dy[i];
            if(x <= n && x >= 1 && y <= n && y >= 1 && mp[x][y])
            {
                if(vis[x][y]) continue;
                d[x][y] = d[tmp.x][tmp.y]+1;
                vis[x][y] = true;
                node1 tmx;
                tmx.x = x;
                tmx.y = y;
                fp.push(tmx);
            }
        }
    }
}

int main()
{
    while(~scanf("%d %d",&n, &m))
    {
        memset(mp, 0, sizeof(mp));
        for(int i = 1; i <= n; i ++)
        {
            scanf("%s", str[i]+1);
            for(int j = 1; j <= n; j++)
                if(str[i][j] == '.') mp[i][j] = 1;
        }
        for(int i = 1; i <= m; i++) scanf("%d %d %d %d",&f[i].x0, &f[i].y0, &f[i].x1, &f[i].y1);
        for(int i = 1; i <= m; i++)
        {
            spfa(f[i].x1, f[i].y1);
            for(int j = 1; j <= m; j++)
            {
                if(i == j)
                {
                    dis[i][j] = 0;
                    continue;
                }
                dis[i][j] = d[f[j].x0][f[j].y0];
            }
        }
        memset(dp, INF, sizeof(dp));
        for(int i = 1; i <= m; i++) dp[1<<(i-1)][i] = 0;
        for(int i = 2; i <= m; i++)
        {
            for(int j = 0; j < (1<<m); j++)
            {
                int sum = 0;
                for(int k = 0; k < m; k++)
                    if(j&(1<<k)) sum++;
                if(sum != i-1) continue;
                for(int k = 0; k < m; k++)
                {
                    if((j&(1<<k)))continue;
                    for(int tt = 1; tt <= m; tt++)
                    {
                        int ans = dp[j][tt]+dis[tt][k+1];
                        dp[j|(1<<k)][k+1] = min(dp[j|(1<<k)][k+1], ans);
                    }
                }
            }
        }
        int ans = INF;
        for(int i = 1; i <= m; i++) ans = min(ans, dp[(1<<m)-1][i]);
        if(ans == INF)
        {
            puts("-1");
            continue;
        }
        printf("%d\n",ans);

    }
    return 0;
}