HDU 1078 FatMouse and Cheese (记忆化搜索)

题意：给定一个n*n的矩阵，问从（0，0）开始走，每次最多水平或者垂直走k个格子，且要保证每次到达的格子要大于前一个，问最大和是多少。

析：一个很简单的记忆搜索，dp[i][j]，表示到达（i，j）的最大和是多少，我们可以反着推出答案。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 10;
const int mod = 1000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn][maxn];
int k;
int dp[maxn][maxn];
bool vis[maxn][maxn];

int dfs(int r, int c){
  int &ans = dp[r][c];
  if(vis[r][c])  return ans;
  vis[r][c] = true;
  for(int i = 1; i <= k; ++i)
    for(int j = 0; j < 4; ++j){
      int x = r + dr[j] * i;
      int y = c + dc[j] * i;
      if(!is_in(x, y) || a[r][c] >= a[x][y])  continue;
      ans = max(ans, dfs(x, y));
    }
  return ans += a[r][c];
}

int main(){
  while(scanf("%d %d", &n, &k) == 2 && k+n != -2){
    m = n;
    for(int i = 0; i < n; ++i)
      for(int j = 0; j < m; ++j)
        scanf("%d", &a[i][j]);
    memset(vis, 0, sizeof vis);
    memset(dp, 0, sizeof dp);
    printf("%d\n", dfs(0, 0));
  }
  return 0;
}