hdu 1078 FatMouse and Cheese【dp】

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1078
题意：每次只能走横着或竖着的 1~k 个格子，求最多能吃到的奶酪。
代码：

#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <assert.h>using namespace std;int n, k;int a[110][110], dp[110][110];int dir[4][2] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };bool is_ok(int x, int y){    if (x < 0 || x >= n || y < 0 || y >= n) return false;    return true;}int res(int x, int y){    int sum = 0,tmp = 0;    if (!dp[x][y])    {        for (int i = 1; i <= k; i++)        {            for (int j = 0; j < 4; j++)            {                int xx = x + dir[j][0] * i;                int yy = y + dir[j][1] * i;                if (is_ok(xx,yy) && a[xx][yy] > a[x][y])                {                    sum = res(xx, yy);                    tmp = max(tmp, sum);                }            }        }        dp[x][y] = tmp + a[x][y];    }    return dp[x][y];}int main(){    while (scanf("%d %d", &n,&k) != EOF)    {        if (n == -1 && k == -1) break;        for (int i = 0; i < n; i++)            for (int j = 0; j < n; j++)                scanf("%d",&a[i][j]);        memset(dp, 0, sizeof(dp));        printf("%d\n", res(0,0));    }    return 0;}

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