POJ 2704 Pascal's Travels 【DFS记忆化搜索】

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Pascal‘s Travels

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5535		Accepted: 2500

Description

An n x n game board is populated with integers, one nonnegative integer per square. The goal is to travel along any legitimate path from the upper left corner to the lower right corner of the board. The integer in any one square dictates how large a step away from that location must be. If the step size would advance travel off the game board, then a step in that particular direction is forbidden. All steps must be either to the right or toward the bottom. Note that a 0 is a dead end which prevents any further progress.

Consider the 4 x 4 board shown in Figure 1, where the solid circle identifies the start position and the dashed circle identifies the target. Figure 2 shows the three paths from the start to the target, with the irrelevant numbers in each removed.


Figure 1	Figure 2

Input

The input contains data for one to thirty boards, followed by a final line containing only the integer -1. The data for a board starts with a line containing a single positive integer n, 4 <= n <= 34, which is the number of rows in this board. This is followed by n rows of data. Each row contains n single digits, 0-9, with no spaces between them.

Output

The output consists of one line for each board, containing a single integer, which is the number of paths from the upper left corner to the lower right corner. There will be fewer than 2⁶³ paths for any board.

Sample Input

Sample Output

3
0
7

Hint

Brute force methods examining every path will likely exceed the allotted time limit. 64-bit integer values are available as long values in Java or long long values using the contest‘s C/C++ compilers.

Source

Mid-Central USA 2005

题意概括：

有一个N*N的游戏板，每一格的数字代表可以跳的步数，起点在左下角，每次可以选择向下或者向右跳，问从左上角起点（固定）跳到右下角终点（固定）的路径有几条。

解题思路：

DFS模拟暴力跳的可能性，记忆化搜索需要记录从当前格可以到达终点的路径数。

！！！因为每天路径都是独一无二的，需要一个标记数组 vis （一开始想着每次都是向下向右应该不会重复，不过wa掉了）

AC code：

 1 ///POJ 2704 记忆化搜索
 2 #include <cmath>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <iostream>
 6 #include <algorithm>
 7 #define INF 0x3f3f3f3f
 8 #define ll long long int
 9 using namespace std;
10 const int MAXN = 35;
11 
12 ll d[MAXN][MAXN];
13 int mmp[MAXN][MAXN];
14 bool vis[MAXN][MAXN];
15 int N;
16 
17 bool ok(int x, int y)
18 {
19     if(x >= 1 && x <= N && y >= 1 && y <= N) return true;
20     else return false;
21 }
22 ll dfs(int x, int y)
23 {
24     if(d[x][y] || (x==N && y==N)) return d[x][y];
25     if(!ok(x+mmp[x][y], y) && !ok(x, y+mmp[x][y])) d[x][y] = -1;
26     if(ok(x+mmp[x][y], y) && d[x+mmp[x][y]][y]!=-1)
27     {
28         if(!vis[x+mmp[x][y]][y])
29         {
30             vis[x+mmp[x][y]][y] = 1;
31             ll len = dfs(x+mmp[x][y], y);
32             vis[x+mmp[x][y]][y] = 0;
33             if(len > 0) d[x][y] += len;
34         }
35     }
36     if(ok(x, y+mmp[x][y]) && d[x][y+mmp[x][y]]!=-1)
37     {
38         if(!vis[x][y+mmp[x][y]])
39         {
40             vis[x][y+mmp[x][y]] = 1;
41             ll len = dfs(x, y+mmp[x][y]);
42             vis[x][y+mmp[x][y]] = 0;
43             if(len > 0) d[x][y] += len;
44         }
45     }
46     return d[x][y];
47 }
48 int main()
49 {
50     int T = 30;
51     char str[MAXN][MAXN];
52     while(T--)
53     {
54         scanf("%d", &N);
55         if(N == -1) break;
56         for(int i = 1; i <= N; i++)
57                 scanf("%s", &str[i]);
58         for(int i = 1; i <= N; i++)
59             for(int j = 0; j < N; j++)
60             mmp[i][j+1] = str[i][j]-‘0‘;
61         memset(d, 0, sizeof(d));
62         memset(vis, 0, sizeof(vis));
63         d[N][N] = 1;
64         ll ans = dfs(1, 1);
65         printf("%lld\n", ans);
66     }
67     return 0;
68 }

POJ 2704 Pascal's Travels 【DFS记忆化搜索】

原文：https://www.cnblogs.com/ymzjj/p/9495480.html

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