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# hdoj 1028/poj 2704 Pascal's Travels(记忆化搜索||dp)

有个小球，只能向右边或下边滚动，而且它下一步滚动的步数是它在当前点上的数字，如果是0表示进入一个死胡同。求它从左上角到右下角到路径数目。

```#include <stdio.h>
#include <string.h>
#define ll __int64
int n;
ll vis[36][36];
char board[36][36];

ll dfs(int x,int y)
{
if(x==n-1&&y==n-1)
return 1;
if(board[x][y]=='0')
return 0;
if(vis[x][y])
return vis[x][y];
if(x+board[x][y]-'0'<n)
vis[x][y]+=dfs(x+board[x][y]-'0',y);
if(y+board[x][y]-'0'<n)
vis[x][y]+=dfs(x,y+board[x][y]-'0');
return vis[x][y];
}

int main()
{
while (scanf("%d",&n)!=EOF)
{
if(n == -1)
break;
for (int i = 0; i < n; i++)
scanf("%s",board[i]);
memset(vis, 0, sizeof(vis));
printf("%I64d\n", dfs(0,0));
}
return 0;
}
```

dp方法

```#include <stdio.h>
#include <string.h>
#define ll __int64

ll vis[50][50];
char board[50][50];

int main()
{
int i,j,n;
while (scanf("%d",&n)!=EOF)
{
if(n == -1)
break;
for (i = 0; i < n; i++)
scanf("%s",board[i]);
memset(vis, 0, sizeof(vis));
vis[0][0] = 1;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
if(board[i][j] == '0')
continue;
vis[i+board[i][j]-'0'][j] += vis[i][j];
vis[i][j+board[i][j]-'0'] += vis[i][j];
}
}
printf("%I64d\n", vis[n-1][n-1]);
}
return 0;
}
```