hdu5433 Xiao Ming climbing（BestCoder Round #55 ($) ）

Xiao Ming climbing

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 489 Accepted Submission(s): 120

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
   
   
   
   

Source

BestCoder Round #55 ($) 

题意：给你一个图，求起点到终点花费的最小体力，体力的计算公式为 (abs(H1−H2))/k，k为当前斗志，每走一步，斗志减1.

分析：普通的bfs，只是注意每一个点可能多次遍历，要找到该点花费最小的体力，最后在这些点内找出最终结果（详解见代码）。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

int n,m,k;
int a,b,c,d;
double minx;
char ch[55];
int mat[55][55];
double vis[55][55][55];//vis[i][j][k]表示在（i，j）这个坐标的时候所剩k斗志的体力值
int dir[4][2]={{0,-1},{0,1},{-1,0},{1,0}};

struct node
{
    int x,y,t;//t为斗志
    double ans;//体力
}s[100005];

void bfs(node st)
{
    queue<node> q;
    node now, next;
    st.t=k;
    st.ans=0.0;
    vis[st.x][st.y][st.t]=0;
    q.push(st);
    while (!q.empty())
    {
        now = q.front();
        if (now.t<=0) return ;
        for (int i=0; i<4; i++)//平常的bfs
        {
            next.x = now.x+dir[i][0];
            next.y = now.y+dir[i][1];
            if (next.x<0||next.x>=n||next.y<0||next.y>=m) continue;
            if (mat[next.x][next.y]==-1) continue;
            
            next.ans = now.ans+(double)abs(mat[now.x][now.y]-mat[next.x][next.y])/(double)now.t;
            next.t = now.t-1;
        
            if (vis[next.x][next.y][next.t]-next.ans>0)//当前的得到的体力值比之前得到的体力值小
            {
                vis[next.x][next.y][next.t]=next.ans;//更新
                q.push(next);//入队
            }
        }
        q.pop();
    }
    return ;
}

int main ()
{
    int T;
    scanf ("%d",&T);
    while (T--)
    {
        scanf ("%d%d%d",&n,&m,&k);
        for (int i=0; i<n; i++)
        {
            scanf ("%s",ch);
            for (int j=0; j<m; j++)
            {
                if (ch[j] == '#')
                    mat[i][j] = -1;
                else
                    mat[i][j] = ch[j]-'0';
            }
        }
        scanf ("%d%d",&a,&b);
        scanf ("%d%d",&c,&d);
        if(k<=0)
        {
            printf("No Answer\n");
            continue;
        }
        a--,b--,c--,d--;
        node st;
        st.x=a;st.y=b;
        //CL(vis);
        for (int i=0; i<=n; i++)//初始化
            for (int j=0; j<=m; j++)
                for (int r=0; r<=k; r++)
                    vis[i][j][r]=INF;
        bfs(st);
        minx=vis[c][d][1];
        for (int i=1; i<=k; i++)//最后扫一遍，得到最小体力
            minx=min(minx, vis[c][d][i]);
        if (minx-INF<0)
            printf ("%.2lf\n",minx);
        else
            printf ("No Answer\n");
    }
    return 0;
}