HDU 5433 Xiao Ming climbing 动态规划

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5433

Xiao Ming climbing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1346    Accepted Submission(s): 384

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

题解：

　　看网上都是bfs的解法，这里来一发动态规划。

　　设dp[i][j][k]代表小明走到(i,j)时还剩k个单位的fighting will的状态；

　　令(i',j') 表示（i，j）上下左右的某一点，那么易得转移方程：

　　　　dp[i][j][k]=min(dp[i][j][k],dp[i'][j'][k+1]+abs(H[i][j]-H[i'][j'])/(k+1))

　　由于状态转移的顺序比较复杂，所有可以用记忆化搜索的方式来求解。

　　最终ans=min(dp[x2][y2][1],......,dp[x2][y2][k]]).

代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;

 const int maxn=;

 double dp[maxn][maxn][maxn];
 bool vis[maxn][maxn][maxn];
 char mat[maxn][maxn];

 int n,m,len;
 int X1,Y1,X2,Y2;

 void init(){
     memset(vis,,sizeof(vis));
     memset(dp,0x7f,sizeof(dp));
 }

 const int dx[]={-,,,};
 const int dy[]={,,-,};
 double solve(int x,int y,int k){
     if(vis[x][y][k]) return dp[x][y][k];
     vis[x][y][k]=;
     for(int i=;i<;i++){
         int tx=x+dx[i],ty=y+dy[i];
         if(tx<||tx>n||ty<||ty>m||k+>len||mat[tx][ty]=='#') continue;
         double add=fabs((mat[x][y]-mat[tx][ty])*1.0)/(k+);
         dp[x][y][k]=min(dp[x][y][k],solve(tx,ty,k+)+add);
     }
     return dp[x][y][k];
 }

 int main(){
     int tc;
     scanf("%d",&tc);
     while(tc--){
         init();
         scanf("%d%d%d",&n,&m,&len);
         for(int i=;i<=n;i++) scanf("%s",mat[i]+);
         scanf("%d%d%d%d",&X1,&Y1,&X2,&Y2);
         dp[X1][Y1][len]=; vis[X1][Y1][len]=;
         double ans=0x3f;
         int flag=;
         for(int k=len;k>=;k--){
             double tmp=solve(X2,Y2,k);
             if(ans>tmp){
                 flag=;
                 ans=tmp;
             }
         }
         if(flag) printf("%.2lf\n",ans);
         else printf("No Answer\n");
     }
     return ;
 }