POJ1088 滑雪题解+HDU 1078（记忆化搜索DP)

Description

Michael喜欢滑雪百这并不奇怪，因为滑雪的确很刺激。可是为了获得速度，滑的区域必须向下倾斜，而且当你滑到坡底，你不得不再次走上坡或者等待升降机来载你。Michael想知道载一个区域中最长底滑坡。区域由一个二维数组给出。数组的每个数字代表点的高度。下面是一个例子 1 2 3 4 5

16 17 18 19 6

15 24 25 20 7

14 23 22 21 8

13 12 11 10 9

一个人可以从某个点滑向上下左右相邻四个点之一，当且仅当高度减小。在上面的例子中，一条可滑行的滑坡为24-17-16-1。当然25-24-23-…-3-2-1更长。事实上，这是最长的一条。 Input

输入的第一行表示区域的行数R和列数C(1 <= R,C <= 100)。下面是R行，每行有C个整数，代表高度h，0<=h<=10000。 Output

输出最长区域的长度。 Sample Input

5 5 1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9 Sample Output

#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
#define mst(a,b)  memset((a),(b),sizeof(a));
const int max1=105;
long long int dp[max1][max1];
//int sum[max1];
long long int ob[max1][max1];
int dx[5]={0,-1,0,1,0},
    dy[5]={0,0,-1,0,1};
    //dx与dy一组，构成上下左右四个方向
long long int r,//长
            c,//宽
            t,//做中间变量，用来寻找最大的ans
            ans;//预设答案变量
int search(int x,int y);//声明search函数
int main()
{
    cin>>r>>c;//输入长宽
    ans=0;//无合适条件下设ans为最小值0
    for(int i=1;i<=r;i++)
        for(int j=1;j<=c;j++)
            cin>>ob[i][j];//输入矩阵
    for(int i=1;i<=r;i++)
        for(int j=1;j<=c;j++){//遍历矩阵中每一个点
            t=search(i,j);//调用searc函数
            dp[i][j]=t;
            if(t>ans) ans=t;
        }
        cout<<ans<<endl;
}
int search(int x,int y)
{
    int w,tmp,nx,ny;
    if(dp[x][y]>0){ //如果dp大于0代表dp被调用过可以直接使用
        return(dp[x][y]);//返回dp[i][j]的值
    }
    w=1;
    for(int i=1;i<=4;i++){//遍历四个方向
        nx=x+dx[i];
        ny=y+dy[i];
        if((nx>=1)&&(nx<=r)&&(ny>=1)&&(ny<=c)&&(ob[x][y]<ob[nx][ny])){//边界
            tmp=search(nx,ny)+1;//递归求当前点能够到达的最大值
            if(tmp>w) w=tmp;
        }
    }
        dp[x][y]=w;
        return w;
}
/*
5 5
1  2  3  4  5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
*/

FatMouse and Cheese

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

#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
int ob[10050][105];
int dp[10050][105];
int dx[4]={0,1,0,-1},
    dy[4]={1,0,-1,0};
int x,y,ans,temp,k;
int search(int q,int w);
int main()
{
    while(cin>>x>>k)
    {
        memset(ob,0,sizeof(ob));
        memset(dp,0,sizeof(dp));
        if(x==-1||y==-1) break;
        y=x;
        ans=0;
        for(int i=1;i<=y;i++)
            for(int t=1;t<=x;t++) cin>>ob[i][t];
            cout<<search(1,1)<<endl;
    }
}
int search(int q,int w)
{
    int nx,ny,t=0;
    if(dp[q][w]>0) return dp[q][w];
    for(int j=1;j<=k;j++)
    for(int i=0;i<4;i++)
    {
        nx=q+dx[i]*j;
        ny=w+dy[i]*j;
        if((nx>=1)&&(nx<=x)&&(ny>=1)&&(ny<=y)&&(ob[q][w]<ob[nx][ny]))
            t=max(t,search(nx,ny));
    }
    return dp[q][w]=t+ob[q][w];
}