Description
The cows have revised their game of leapcow. They now play in the middle of a huge pasture upon which they have marked a grid that bears a remarkable resemblance to a chessboard of N rows and N columns (3 <= N <= 365).
Here‘s how they set up the board for the new leapcow game:
* First, the cows obtain N x N squares of paper. They write the integers from 1 through N x N, one number on each piece of paper.
* Second, the ‘number cow‘ places the papers on the N x N squares in an order of her choosing.
Each of the remaining cows then tries to maximize her score in the game.
* First, she chooses a starting square and notes its number.
* Then, she makes a ‘knight‘ move (like the knight on a chess board) to a square with a higher number. If she‘s particularly strong, she leaps to the that square; otherwise she walks.
* She continues to make ‘knight‘ moves to higher numbered squares until no more moves are possible.
Each square visited by the ‘knight‘ earns the competitor a single point. The cow with the most points wins the game.
Help the cows figure out the best possible way to play the game.
Input
* Line 1: A single integer: the size of the board
* Lines 2.. ...: These lines contain space-separated integers that tell the contents of the chessboard. The first set of lines (starting at the second line of the input file) represents the first row on the chessboard; the next set of lines represents the next row, and so on. To keep the input lines of reasonable length, when N > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its own line.
Output
* Line 1: A single integer that is the winning cow‘s score; call it W.
* Lines 2..W+1: Output, one per line, the integers that are the starting square, the next square the winning cow visits, and so on through the last square. If a winning cow can choose more than one path, show the path that would be the ‘smallest‘ if the paths were sorted by comparing their respective ‘square numbers‘.
Sample Input
4 1 3 2 16 4 10 6 7 8 11 5 12 9 13 14 15
Sample Output
7 2 4 5 9 10 12 13 题意：给你一个矩阵，问你按照象棋马的走法，下一步比上一步的数大，问长度最长的序列是多长，然后输出序列。如果有多个最长序列输出字典序最小的那个。这是看到的一个代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <queue>
5 #include <vector>
6 #include <algorithm>
7 #include <map>
8 using namespace std;
9 typedef pair<int,int>P;
10 const double eps=1e-9;
11 const int maxn=200100;
12 const int mod=1e9+7;
13 const int INF=1e9;
14 int M[370][370],dp[370][370];
15 int N;
16 int dx[8]= {1,1,2,2,-1,-1,-2,-2};
17 int dy[8]= {2,-2,1,-1,2,-2,1,-1};
18 P path[370][370];
19 vector<P>res;
20 int DP(int x,int y)
21 {
22 int &m=dp[x][y];
23 if(m) return m;
24 m=1;
25 for(int i=0; i<8; i++)
26 {
27 int nx=x+dx[i];
28 int ny=y+dy[i];
29 if(1<=nx&&nx<=N&&1<=ny&&ny<=N&&M[nx][ny]>M[x][y])
30 {
31 if(m<DP(nx,ny)+1)
32 {
33 m=DP(nx,ny)+1;
34 path[x][y]=make_pair(nx,ny);
35 }
36 else if(DP(nx,ny)+1==m)
37 {
38 if(M[path[x][y].first][path[x][y].second]>M[nx][ny])
39 {
40 path[x][y]=make_pair(nx,ny);
41 }
42 }
43 }
44 }
45 return m;
46 }
47 int main()
48 {
49 while(~scanf("%d",&N))
50 {
51 for(int i=1; i<=N; i++)
52 {
53 for(int j=1; j<=N; j++)
54 {
55 scanf("%d",&M[i][j]);
56 }
57 }
58 int MAX=1,cnt;
59 for(int i=1; i<=N; i++)
60 {
61 for(int j=1; j<=N; j++)
62 {
63 cnt=DP(i,j);
64 if(MAX<cnt)
65 {
66 MAX=cnt;
67 res.clear();
68 res.push_back(make_pair(i,j));
69 }
70 else if(MAX==cnt)
71 {
72 res.push_back(make_pair(i,j));
73 }
74 }
75 }
76 int keyx,keyy,key;
77 key=INF;
78 for(int i=0; i<res.size(); i++)
79 if(M[res[i].first][res[i].second]<key)
80 {
81 key=M[res[i].first][res[i].second];
82 keyx=res[i].first;
83 keyy=res[i].second;
84 }
85 printf("%d\n",MAX);
86 while(1)
87 {
88 printf("%d\n",M[keyx][keyy]);
89 int t=path[keyx][keyy].second;
90 keyx=path[keyx][keyy].first;
91 keyy=t;
92 if(!keyx)
93 break;
94 }
95 }
96 return 0;
97 }
知识点：
这个代码的主要想法是，如同最短路一样，path中储存的是当前节点的下一步应该走的位置，然后进行搜索直到遍历了所有的点。