程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> 關於C語言 >> hdu 1078 FatMouse and Cheese(記憶化搜索)

hdu 1078 FatMouse and Cheese(記憶化搜索)

編輯:關於C語言

hdu 1078 FatMouse and Cheese(記憶化搜索)


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

37
Source   Zhejiang University Training Contest 2001
Recommend   We have carefully selected several similar problems for you:10801074102410811025
Statistic|Submit|Discuss|Note

普通搜索會超時的 感覺這道題和滑雪(poj1088,nyoj10)一樣 只不過多了一些方向

首先附上超時代碼 +理解錯誤(He eats up the cheese where he stands and then runs either horizontally or vertically to another location.)

#include 
#include 
int n,k;
int map[105][105];
bool vis[105][105];
int dir[4][2]={1,0,-1,0,0,1,0,-1};
int max;
bool limit(int x1,int y1,int x,int y)
{
    if(vis[x1][y1]||x1<0||y1<0||x1>=n||y1>=n||map[x1][y1]<=map[x][y])
    return false;
    return true;
}
void dfs(int x,int y,int sum)
{
    if(sum>max)
    max=sum;
    for(int i=0;in+n)
        k=n+n;
        for(int i=0;i理解為可以任意走k步了 Wa也是應該的

  

記憶化搜索:

 

#include 
#include 
#include 
#include 
using namespace std;
int n,k,result;
int map[105][105];
int dp[105][105];
int dir[4][2]={1,0,-1,0,0,1,0,-1};
struct node
{
	int x,y;
};
bool limit(int x1,int y1,int x,int y)
{
    if(x1<0||y1<0||x1>=n||y1>=n||map[x1][y1]<=map[x][y])
    return false;
    return true;
}
int dfs(int x,int y)
{
	if(dp[x][y]) 
	return dp[x][y];
	for(int i=1;i<=k;i++)
	{
		for(int j=0;j<4;j++)
		{
			
			int x1=x+dir[j][0]*i;
			int y1=y+dir[j][1]*i;
			if(limit(x1,y1,x,y))
			{
				int z=dfs(x1,y1);
				if(dp[x][y]

 

  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved