标签：for before current ali out names fir 左右 led
http://acm.hdu.edu.cn/showproblem.php?pid=1078
FatMouse and Cheese
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17910 Accepted Submission(s): 7619
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
题意：从（0，0）出发每次上下左右走一步，每次可以走1-k次停下吃奶酪，求最多可以吃多少奶酪。
//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF 0x3f3f3f3f
#define mod 998244353
#define PI acos(-1)
using namespace std;
typedef long long ll ;
int a[109][109] ;
int dp[109][109];
int dir[4][2] = {{1 , 0} , {-1 , 0} , {0 , 1} , {0 , -1}};
int n , k ;
int dfs(int x , int y)
{
if(dp[x][y])//避免重复赋值，减少时间
return dp[x][y];
dp[x][y] = a[x][y] ;
for(int i = 0 ; i < 4 ; i++)
{
for(int j = 1 ; j <= k ; j++)
{
int nx = x + dir[i][0]*j ;
int ny = y + dir[i][1]*j ;
if(nx >= 0 && nx < n && ny >= 0 && ny < n)
{
if(a[nx][ny] > a[x][y])
{
dp[x][y] = max(dp[x][y] , dfs(nx , ny) + a[x][y]);
}
}
}
}
return dp[x][y] ;
}
int main()
{
while(~scanf("%d%d" , &n , &k) && (n != -1 || k != -1))
{
memset(dp , 0 , sizeof(dp));
for(int i = 0 ; i < n ; i++)
{
for(int j = 0 ; j < n ; j++)
{
scanf("%d" , &a[i][j]);
}
}
cout << dfs(0 , 0) << endl ;
}
return 0 ;
}
dp（动态规划之最佳路径+dfs）
标签：for before current ali out names fir 左右 led
原文地址：https://www.cnblogs.com/nonames/p/11779711.html
