一道记忆搜索题，记忆搜索题就是搜索的形式+DP的思想！
题目：
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
思想：
一道记忆搜索题，记忆搜索题就是搜索的形式+DP的思想！
代码：
超时代码：
1 #include<stdio.h>
2 #include<string.h>
3 //#include<stdlib.h>
4 /*#include<iostream>
5 #include<algorithm>
6 using namespace std;*/
7
8 int getmax(int x, int y){
9 return x > y ? x : y;
10 }
11
12 typedef struct{
13 int x;
14 int y;
15 }List;
16 List list[10001];
17
18 int dir[4][2]={0, -1, -1, 0, 0, 1, 1, 0};
19
20
21 int main(){
22 int n, k, block[101][101], i, j, ii, jj, dp[101][101], max, lastmax, startx, starty, start, num;
23 while(scanf("%d %d", &n, &k) && (n != -1 || k != -1)){
24 max = -1;
25 for(i = 0; i < n; i ++){
26 for(j = 0; j < n; j ++){
27 scanf("%d", &block[i][j]);
28 max = getmax(max, block[i][j]);
29 }
30 }
31 memset(dp, -1, sizeof(dp));
32 dp[0][0] = block[0][0];
33 start = dp[0][0];
34 startx = 0;
35 starty = 0;
36 num = 0;
37 lastmax = dp[0][0];
38 do{
39 for(i = 0; i < 4; i ++){
40 if(block[startx][starty] == max){
41 break;
42 }
43 for(j = 1; j <= k; j ++){
44 ii = startx + dir[i][0] * j;
45 jj = starty + dir[i][1] * j;
46 if(ii < 0 || jj < 0 || ii >= n || jj >= n || block[ii][jj] <= block[startx][starty]){
47 continue;
48 }
49 dp[ii][jj] = getmax(dp[ii][jj], start + block[ii][jj]);
50 lastmax = getmax(dp[ii][jj], lastmax);
51 list[num].x = ii;
52 list[num ++].y = jj;
53 }
54 }
55 if(num == 0){
56 num --;
57 continue;
58 }
59 startx = list[-- num].x;
60 starty = list[num].y;
61 start = dp[startx][starty];
62 }while(num != -1);
63 printf("%d\n", lastmax);
64 }
65 return 0;
66 }
奋进（超时代码）
望大神指点，个人觉得跟下面的正确代码时间复杂度差不多啊！
正确代码：
1 #include<stdio.h>
2 #include<string.h>
3
4 int map[105][105];
5 int value[105][105];
6 int dir[4][2]={0,1,0,-1,1,0,-1,0};
7 int n,k;
8
9 int dfs(int sx,int sy)
10 {
11 int i,j,x,y,max=0,temp;
12 if(value[sx][sy]!=-1) //该点已经计算过，直接返回值
13 return value[sx][sy];
14 for(i=0;i<4;i++) //2个循环控制该点所能走的地方
15 for(j=1;j<=k;j++)
16 {
17 x=sx+dir[i][0]*j;
18 y=sy+dir[i][1]*j;
19 if(x<0||x>=n||y<0||y>=n||map[x][y]<=map[sx][sy]) //剪枝
20 continue;
21 temp=dfs(x,y);
22 if(temp>max) //max保存由该点所能到达之处能吃到的最大奶酪数
23
24 max=temp;
25 }
26 value[sx][sy]=map[sx][sy]+max; //得到由该点出发所能吃到的最大奶酪数，保存在value里
27 return value[sx][sy];
28 }
29
30 int main()
31 {
32 int i,j;
33 while(scanf("%d%d",&n,&k)!=EOF)
34 {
35 if(n==-1&&k==-1)
36 break;
37 memset(value,-1,sizeof(value)); //初始化数组(不能为0)
38 for(i=0;i<n;i++)
39 for(j=0;j<n;j++)
40 scanf("%d",&map[i][j]);
41 printf("%d\n",dfs(0,0));
42 }
43 return 0;
44 }
网上大神代码
总结：
尽量不要用do....while();