HDU 1078

 这题之后的题我都不会了，吃饭前还感觉DP意外的很简单啊，吃完饭回来就凉凉了，百度了kaungbin大神的代码，看了之后总算是会写了，但是关于MAX为何要初始化为S[0][0]还是想了挺久的，最后想到s[0][0]可能直接就是最大的，不能往其他地方转移，这样的话就不能更新MAX了，总算是AC了

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. 

InputThere 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

代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[105][105];
int s[105][105];
struct R{
	int x,y;
	int v;
};
R a[105*105];
int cmp(R a,R b){
	return a.v<b.v;
}
int main(){
	int n,d;
	while(scanf("%d%d",&n,&d)){
		if(n==-1&&d==-1) break;
		int Max;
		int cnt=0;
		memset(dp,-1,sizeof(dp));
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				scanf("%d",&s[i][j]);
				if(i!=0||j!=0){
					a[cnt].x=j;
					a[cnt].y=i;
					a[cnt].v=s[i][j];
					cnt++;
				}
			}
		}
		sort(a,a+cnt,cmp);
		dp[0][0]=s[0][0];
		Max=s[0][0]; //因为存在s[0][0]为最大的可能，此时不会进行状态的转移，故要给MAx初始化为s[0][0] 
		for(int i=0;i<cnt;i++){
			int x=a[i].x;
			int y=a[i].y;
			for(int xx=max(0,x-d);xx<=min(n-1,x+d);xx++){
				if(s[y][xx]>=s[y][x]) continue;
				if(dp[y][xx]==-1) continue;
				dp[y][x]=max(dp[y][x],dp[y][xx]+s[y][x]);
			}
			for(int yy=max(0,y-d);yy<=min(n-1,y+d);yy++){
				if(s[yy][x]>=s[y][x]) continue;
				if(dp[yy][x]==-1) continue;
				dp[y][x]=max(dp[y][x],dp[yy][x]+s[y][x]);
			}
			Max=max(Max,dp[y][x]);
		}
		printf("%d\n",Max);
	}
}