D - FatMouse and Cheese <HDU 1078>
阿新 • • 發佈：2018-07-07
mil bits ace tput 初始化 number pac its 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.
D - FatMouse and Cheese
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.
OutputFor 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
37
題意:給一個圖 從(0,0) 可以垂直或水平走 每次可以走小於等於k的任意距離 但每步的值 必須大於前一步的值 在無路可走時結束(周圍k範圍內無比該步大的值) 求可走完的最大值
思路:用dfs跑圖 在dfs中設置cnt 用來記錄上一個狀態的最大值 將cnt+當前圖裏的值記錄在二維dp數組中
#include<bits/stdc++.h> using namespace std; int n,k,maps[101][101],dp[101][101]; int dix[4]={0,1,0,-1};//x移動數組 int diy[4]={1,0,-1,0};//y移動數組 int dfs(int x,int y) { if(dp[x][y]) return dp[x][y];//當dp數組不為0 是證明 x，y走過了 int cnt=0;//用於記錄之前值 for(int i=0;i<4;i++) { for(int j=1;j<=k;j++) { int dx=x+dix[i]*j; int dy=y+diy[i]*j; if(dx<0||dx>=n||dy<0||dy>=n)//判斷越界的條件 break; if(maps[x][y]<maps[dx][dy]) cnt=max(cnt,dfs(dx,dy));//更新cnt } } return dp[x][y]=cnt+maps[x][y]; } int main() { while(1) { cin>>n>>k; memset(dp,0,sizeof(dp));//初始化 if(n==-1&&k==-1) break; for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { cin>>maps[i][j]; } } cout<<dfs(0,0)<<endl; } return 0; }
D - FatMouse and Cheese <HDU 1078>