记忆化搜索

 

  记忆化搜索是基于 dfs搜索的一种改进 

是当 搜索到已经搜索过的位置时 通过标记 来进行剪枝的方法

例题

 

A - Longest Run on a Snowboard

 

Michael likes snowboarding. That’s not very surprising, since snowboarding is really great. The bad thing is that in order to gain speed, the area must slide downwards. Another disadvantage is that when you’ve reached the bottom of the hill you have to walk up again or wait for the ski-lift. Michael would like to know how long the longest run in an area is. That area is given by a grid of numbers, defining the heights at those points. Look at this example:
1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9
One can slide down from one point to a connected other one if and only if the height decreases. One point is connected to another if it’s at left, at right, above or below it. In the sample map, a possible slide would be 24-17-16-1 (start at 24, end at 1). Of course if you would go 25-24-23-...-3-2-1, it would be a much longer run. In fact, it’s the longest possible.
Input
The first line contains the number of test cases N. Each test case starts with a line containing the name (it’s a single string), the number of rows R and the number of columns C. After that follow R lines with C numbers each, defining the heights. R and C won’t be bigger than 100, N not bigger than 15 and the heights are always in the range from 0 to 100.
Output
For each test case, print a line containing the name of the area, a colon, a space and the length of the longest run one can slide down in that area.
 

 

该题 就可以通过记忆化搜索通过

#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;    
char s[10000];
int n,m,t,maxx;
int d[200][200],dp[200][200];
int dx[5]={1,0,-1,0};
int dy[5]={0,1,0,-1}; 
int dfs(int x,int y){

    if(dp[x][y]!=-1)
       return dp[x][y];
    else {
	int maxx=1;
    for(int i=0;i<4;i++){
    	  int a=x+dx[i];
    	  int b=y+dy[i];
    	  if(a>=0&&b>=0&&a<n&&b<m){
    	  	    if(d[x][y]>d[a][b])
    	  	    	maxx=max(maxx,dfs(a,b)+1);
		  }
    }
    dp[x][y]=maxx;
	return maxx;
}
} 
int main()
{
	scanf("%d",&t);
	while(t--){
		maxx=0;
		scanf("%s",s);
		scanf("%d%d",&n,&m);
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
				scanf("%d",&d[i][j]);
		memset(dp,-1,sizeof(dp));
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
				maxx=max(maxx,dfs(i,j)) ;
		printf("%s: ",s);
		printf("%d\n",maxx);	
	}
	return 0;
}