dp备忘录 pku2704

Total Submissions: 3554

Accepted: 1502

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.



解答：
一道很简单的备忘录dp，最开始想搜索来着，一经仔细思考发现是个dp问题。
思路：从左到右，从上至下，依次计算当前节点(q)可以step到哪里(记为p)，在p的原值基础上加上当前节点q的所有到达路径数。最后在矩阵右下的终点里就记录了最终的路径数
注意：结果是long long int。

#include 
#include 

int board[50][50];
long long count[50][50];
char string[50];

void input(int size);
long long get_path(int size);

int main(int argc, char *argv[])
{
    int n, i;
    while (scanf("%d", &n) && n != -1)
    {
        memset(count, 0, sizeof(count));
        count[0][0] = 1;
        input(n);
        printf("%lld\n", get_path(n));
    }
}

void input(int size)
{
    int i, j;
    char *ch;
    for (i=0 ; i    {
        scanf("%s", string);
        ch = &string[0];
        for (j=0 ; j        {
            board[i][j] = *ch - '0';
            ch++;
        }
    }
}

long long get_path(int size)
{
    int i, j, step;
    for (i=0 ; i    {
        for (j=0 ; j        {
            step = board[i][j];
            if (0 == step)
            {
                continue;
            }
            count[i + step][j] += count[i][j];
            count[i][j + step] += count[i][j];
        }
    }
    return count[size-1][size-1];
}