HDU 4856 tunnels (BFS + pressure DP)

Give you an N * n graph to reach all ". "Point," # "cannot pass. Each group of M has an entrance and an exit. The entrance can be transferred to the exit. I do not know the sequence of passing through the M group, let you find all the "..

Idea: First BFS out the shortest distance between all M, and then DP [J] [I] indicates that the shortest path of step I is started in the J state, enumeration finds the smallest DP [1 <m-1] [I], that is, the shortest distance. Otherwise, the output is "-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 descriptionbob is traveling in Xi'an. he finds eclipsecret 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 ).
Inputthe 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 X1, Y1, X2, Y2, indicating there is a tunnel with entrence in (x1, Y1) and exit in (X2, Y2 ). it's guaranteed that (x1, Y1) and (X2, Y2) in the map are both empty grid.
Outputfor 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 41 2 3 52 3 3 15 4 2 1

 
Sample output

7

 

#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 __int64using 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;}

HDU 4856 tunnels (BFS + pressure DP)