[C++知识库]POJ2704 HDU1208 LA3390 Pascal‘s Travels【递推+记忆化搜索】

Pascal’s Travels
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5656 Accepted: 2538

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

4
2331
1213
1231
3110
4
3332
1213
1232
2120
5
11101
01111
11111
11101
11101
-1

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

问题链接：POJ2704 HDU1208 LA3390 Pascal’s Travels
问题简述：给定一个棋盘，棋盘上的每个数字表示需要走几格，只能向下或向右走。问从左上角走到右下角有几种走法？
问题分析：一种方法是用递推实现，通常称为DP实现；另外一种方法是用DFS实现，用记忆化搜索来加快速度。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* POJ2704 HDU1208 LA3390 Pascal's Travels */

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N = 34;
char b[N][N + 1];
long long dp[N][N];
int n;

int main()
{
    while (~scanf("%d", &n) && n != -1) {
        getchar();
        for (int i = 0; i < n; i++) {
            scanf("%s", b[i]);
            for (int j = 0; j < n; j++)
                b[i][j] -= '0';
        }

        memset(dp, 0, sizeof dp);
        dp[0][0] = 1;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                if (b[i][j]) {
                    if (i + b[i][j] < n)
                        dp[i + b[i][j]][j] += dp[i][j];
                    if (j + b[i][j] < n)
                        dp[i][j + b[i][j]] += dp[i][j];
                }

        printf("%lld\n", dp[n - 1][n - 1]);
    }

    return 0;
}

AC的C++语言程序如下：

/* POJ2704 HDU1208 LA3390 Pascal's Travels */

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int drow[] = {0, 1};
const int dcol[] = {1, 0};
const int N = 34;
char b[N][N + 1];
int vis[N][N];
long long dp[N][N];
int n;

long long dfs(int row, int col)
{
    if (row < 0 || row >= n || col < 0 || col >= n) return 0;
    if (vis[row][col]) return dp[row][col];
    vis[row][col] = 1;
    for (int i = 0; i < 2; i++)
        dp[row][col] += dfs(row + (b[row][col] - '0') * drow[i], col + (b[row][col] - '0') * dcol[i]);
    return dp[row][col];

}

int main()
{
    while (~scanf("%d", &n) && n != -1) {
        getchar();
        for (int i = 0; i < n; i++)
            scanf("%s", b[i]);

        memset(vis, 0, sizeof vis);
        memset(dp, 0, sizeof dp);
        vis[n - 1][n - 1] = 1;
        dp[n - 1][n - 1] = 1;

        printf("%lld\n", dfs(0, 0));
    }

    return 0;
}