HDU 5433 Xiao Ming climbing

Problem Description

Due to the curse made by the devil,Xiao Ming is stranded on a mountain and can hardly escape.

This mountain is pretty strange that its underside is a rectangle which size is  n∗m  and every little part has a special coordinate (x,y) and a height  H .

In order to escape from this mountain,Ming needs to find out the devil and beat it to clean up the curse.

At the biginning Xiao Ming has a fighting will  k ,if it turned to  0  Xiao Ming won't be able to fight with the devil,that means failure.

Ming can go to next position (N,E,S,W) from his current position that time every step, (abs(H1−H2))/k  's physical power is spent,and then it cost  1  point of will.

Because of the devil's strong,Ming has to find a way cost least physical power to defeat the devil.

Can you help Xiao Ming to calculate the least physical power he need to consume.

 

Input

The first line of the input is a single integer  T(T≤10) , indicating the number of testcases. 

Then  T  testcases follow.

The first line contains three integers  n,m,k  ,meaning as in the title (1≤n,m≤50,0≤k≤50) .

Then the  N  ×  M  matrix follows.

In matrix , the integer  H  meaning the height of  (i,j) ,and '#' meaning barrier (Xiao Ming can't come to this) .

Then follow two lines,meaning Xiao Ming's coordinate (x1,y1)  and the devil's coordinate (x2,y2) ,coordinates is not a barrier.

 

Output

For each testcase print a line ,if Xiao Ming can beat devil print the least physical power he need to consume,or output " NoAnswer " otherwise.

(The result should be rounded to 2 decimal places)

 

Sample Input

   
   
   
   

    
    
    
    3
4 4 5
2134
2#23
2#22
2221
1 1
3 3
4 4 7
2134
2#23
2#22
2221
1 1
3 3
4 4 50
2#34
2#23
2#22
2#21
1 1
3 3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1.03
0.00
No Answer
   
   
   
   

广搜
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 105;
int T, n, m, k, bx, by, ex, ey;
int a[maxn][maxn];
int c[4][2] = { 1, 0, -1, 0, 0, 1, 0, -1 };
double f[maxn][maxn][maxn];
char s[maxn];

struct point
{
    int x, y, z;
    point(){};
    point(int x, int y, int z) :x(x), y(y), z(z){};
};

void bfs()
{
    for (int i = 1; i <= n;i++)
        for (int j = 1; j <= m;j++)
            for (int u = 0; u <= k; u++) f[i][j][u] = 0x7FFFFFFF;
    queue<point> p;
    p.push(point(bx, by, 0));
    f[bx][by][0] = 0;
    while (!p.empty())
    {
        point q = p.front();    p.pop();
        if (q.z >= k) continue;
        for (int i = 0; i < 4; i++)
        {
            int x = q.x + c[i][0];
            int y = q.y + c[i][1];
            if (x<1 || x>n || y<1 || y>m || a[x][y] < 0) continue;
            double ans = 1.0*abs(a[x][y] - a[q.x][q.y]) / (k - q.z);
            if (f[x][y][q.z + 1]>f[q.x][q.y][q.z] + ans)
            {
                f[x][y][q.z + 1] = f[q.x][q.y][q.z] + ans;
                p.push(point(x, y, q.z + 1));
            }
        }
    }
    double ans = 0x7FFFFFFF;
    for (int i = k - 1; i >= 0; i--) ans = min(ans, f[ex][ey][i]);
    if (ans + 1e-5 < 0x7FFFFFFF) printf("%.2lf\n", ans);
    else printf("No Answer\n");
}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d%d", &n, &m, &k);
        for (int i = 1; i <= n; i++)
        {
            scanf("%s", s + 1);
            for (int j = 1; j <= m; j++) 
                if (s[j] == '#') a[i][j] = -1; else a[i][j] = s[j] - '0';
        }
        scanf("%d%d%d%d", &bx, &by, &ex, &ey);
        if (k) bfs(); else printf("No Answer\n");
    }
    return 0;
}