Time Limit: 1 s

Description

Handoku is sailing on a lake at the North Pole. The lake can be considered as a two-dimensional square plane containing N × N blocks, which is shown in the form of string containing '*' and '#' on the map.

• * : a normal block;
• # : a block containing pack ice.

Handoku is at (1, 1) initially, and his destination is (

N, N). He can only move to one of the four adjacent blocks. Sailing around pack ice is dangerous and stressful, so he needs power to remain vigilant. That means if he moves from a '*' block to a '#' block or moves from a '#' block to any another block, he needs to consume 1 unit power. In other cases, he can enjoy the scene on his boat without consuming any power.

Now Handoku wants to know how many units power will be consumed at least during his sailing on the lake.

 Input

There are several test cases (no more than 20). For each test case, the first line contains a single integer N (3 ≤ N ≤ 50), denoting the size of the lake. For the following N lines, each line contains a N-length string consisting of '*' and '#', denoting the map of the lake.

 Output

For each test case, output exactly one line containing an integer denoting the answer of the question above.

Sample Input

3
**#
**#
*#*
3
##*
#*#
###
4
**##
#**#
##**
###*

 Sample Output

2
4
0

本題題意：

給你一個n*n的地圖，起點為（1,1），終點為（n*n）。該地圖有“*”和“#”組成。只能做上下左右四個方向。從“*”走到“#”消耗1單位能量，從“#”到其他（不管那個位置為啥）都消耗1單位能量，其他情況不消耗能量，問消耗的最小能量是多少。

解題思路：

用BFS+優先佇列來做。但是要注意的是它的標記陣列，走到的地方消耗的能量如果比下一次走到該地方消耗的能量大，那就要變成下一次走到該地消耗的能量。即標記陣列記得是走到該地消耗的最小能量，不斷重新整理標記陣列的值就好。

#include<iostream>
#include<string>
#include<string.h>
#include<sstream>
#include<queue>
using namespace std;
struct node
{
    int x,y,power;
    friend bool operator<(node a,node b)
    {
        return a.power>b.power;
    }
} s,e;
int n;
char a[51][51];
int vi[51][51];
int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}}; //走的方向
void bfs()
{
    priority_queue<node>q;
    node t,x1;
    vi[0][0]=1;
    if (a[s.x][s.y]=='#')
        s.power+=1;
    q.push(s);
    while(!q.empty())
    {
        t=q.top();
        q.pop();
        for(int i=0; i<4; i++)
        {
            x1.y=t.y+dir[i][1];
            x1.x=t.x+dir[i][0];
            if(x1.x>=0&&x1.x<n&&x1.y>=0&&x1.y<n&&vi[x1.x][x1.y]==0)
            {

                vi[x1.x][x1.y]=1;
                if (a[x1.x][x1.y]=='*')
                {
                    x1.power=t.power;
                }
                else
                {
                    if (a[t.x][t.y]=='#')
                        x1.power=t.power+1;
                    else
                    x1.power=t.power+2;
                }
                if (x1.x==n-1&&x1.y==n-1)
                {
                    if(a[n-1][n-1]=='#')
                        x1.power--;
                    cout<<x1.power<<endl;
                    return ;
                }
                q.push(x1);
            }
        }
    }
}
int main()
{
    while(cin>>n)
    {
        memset(vi,0,sizeof(vi));
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
            {
                cin>>a[i][j];
            }
        s.x=0;
        s.y=0;
        s.power=0;
        bfs();
    }
}