hdu1078 FatMouse and Cheese

FatMouse and Cheese

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 87   Accepted Submission(s) : 44

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

FatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. At each grid location Fatmouse has hid between 0 and 100 blocks of cheese in a hole. Now he's going to enjoy his favorite food.

FatMouse begins by standing at location (0,0). He eats up the cheese where he stands and then runs either horizontally or vertically to another location. The problem is that there is a super Cat named Top Killer sitting near his hole, so each time he can run at most k locations to get into the hole before being caught by Top Killer. What is worse -- after eating up the cheese at one location, FatMouse gets fatter. So in order to gain enough energy for his next run, he has to run to a location which have more blocks of cheese than those that were at the current hole.

Given n, k, and the number of blocks of cheese at each grid location, compute the maximum amount of cheese FatMouse can eat before being unable to move.

Input

There are several test cases. Each test case consists of

a line containing two integers between 1 and 100: n and k
n lines, each with n numbers: the first line contains the number of blocks of cheese at locations (0,0) (0,1) ... (0,n-1); the next line contains the number of blocks of cheese at locations (1,0), (1,1), ... (1,n-1), and so on.
The input ends with a pair of -1's.

Output

For each test case output in a line the single integer giving the number of blocks of cheese collected.

Sample Input

3 1
1 2 5
10 11 6
12 12 7
-1 -1

Sample Output

37

这道题与正常bfs不同的是在于行走时可以走1-k格 并且只能走比上一格大的地方，就把这些地方压入队列里面就好了。

/*#include<iostream>
#include<cstring>
#include<queue>
#include<cstdio>
using namespace std;
int ma[555][555];
int xx,yy;
int kk;
int dp[555][555];
int x1[4]={1,0,0,-1};
int y11[4]={0,1,-1,0};
struct point{
    int x;
    int y;
};
bool check(int x,int y){
    if(ma[x][y]==0||x<1||x>xx||y<1||y>yy)
    return 0;
    return 1;
}
int bfs(int xs,int ys){
    queue<point> lss;
    point tt;
    dp[xs][ys]=ma[xs][ys];
    tt.x=xs;
    tt.y=ys;
    lss.push(tt);
    int ans=-1;
    while(!lss.empty()){
        tt=lss.front();
        lss.pop();
        if(dp[tt.x][tt.y]>ans)
        {
            ans=dp[tt.x][tt.y];
        }
        for(int j=0;j<4;j++){
            for(int i=1;i<=kk;i++)
            {
            int xxx=tt.x+x1[j]*i;
            int yyy=tt.y+y11[j]*i;
            if(xxx>=1&&xxx<=xx&&yyy>=1&&yyy<=xx&&ma[xxx][yyy]>ma[tt.x][tt.y])
            {
                if(dp[xxx][yyy]<ma[xxx][yyy]+dp[tt.x][tt.y])
            {
                dp[xxx][yyy]=ma[xxx][yyy]+dp[tt.x][tt.y];
            point hh;
            hh.x=xxx;
            hh.y=yyy;
                lss.push(hh);
            }
            }    
        }
        }
    }
    return ans;
}
int main(){
    while(scanf("%d%d",&xx,&kk)!=EOF){
    
     if(xx==-1||kk==-1)break;
     memset(dp,0,sizeof(dp));
     int i,j;
     for(i=1;i<=xx;i++)
     {
         for(j=1;j<=xx;j++)
         {
             scanf("%d",&ma[i][j]);
         }
     }
     int ans=bfs(1,1);
     printf("%d\n",ans);
}
return 0;
}*/




#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,k;
int ma[555][555];
int dp[555][555];
int xd[4]={0,0,1,-1};
int yd[4]={1,-1,0,0};
int dfs(int x,int y)
{    int i,j;
int ans=0;
    if(!dp[x][y])
    {
        for(j=1;j<=k;j++)
        for(i=0;i<4;i++)
        {
            int xxx=x+xd[i]*j;
            int yyy=y+yd[i]*j;
            if(xxx<1||xxx>n||yyy>n||yyy<1)
            continue;
            if(ma[xxx][yyy]>ma[x][y])
            ans=max(ans,dfs(xxx,yyy));
        }
        dp[x][y]=ans+ma[x][y];
    }
    return dp[x][y];
}
int main()
{    
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==-1||k==-1)break;
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        scanf("%d",&ma[i][j]);
        memset(dp,0,sizeof(dp));
        int ans=dfs(1,1);
        cout<<ans<<endl;
 
    }
    return 0;
}