hdu 1078 FatMouse and Cheese 记忆化搜索

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原文链接：http://www.cnblogs.com/jhz033/p/5600622.html

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#java

本文探讨了一个关于FatMouse在一个n×n的城市网格中尽可能多地收集奶酪的问题。该问题涉及动态规划算法，考虑到每一步FatMouse需要找到比当前位置有更多的奶酪的位置，并且由于超级猫TopKiller的存在，FatMouse每次移动不能超过k步。

FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

Source

Zhejiang University Training Contest 2001

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+10,M=1e6+10,inf=1e9+10;
int dp[N][N];
int a[N][N];
int n,m;
int xx[4]={1,0,-1,0};
int yy[4]={0,1,0,-1};
void init()
{
    memset(dp,0,sizeof(dp));
}
int check(int x,int y)
{
    if(x<1||x>n||y<1||y>n)
    return 0;
    return 1;
}
int dfs(int x,int y)
{
    if(dp[x][y])
    return dp[x][y];
    int maxx=0;
    for(int j=1;j<=m;j++)
    for(int i=0;i<4;i++)
    {
        int u=x+xx[i]*j;
        int v=y+yy[i]*j;
        if(check(u,v)&&a[u][v]>a[x][y])
        maxx=max(maxx,dfs(u,v));
    }
    return dp[x][y]=maxx+a[x][y];
}
int main()
{
    int x,y,z,i,t;
    int T;
    while(~scanf("%d%d",&n,&m))
    {
        if(n<0&&m<0)
            break;
        init();
        for(i=1;i<=n;i++)
        for(t=1;t<=n;t++)
        scanf("%d",&a[i][t]);
        dfs(1,1);
        int maxx=-inf;
        for(i=1;i<=n;i++)
        for(t=1;t<=n;t++)
        maxx=max(maxx,dp[i][t]);
        printf("%d\n",maxx);
    }
    return 0;
}