题意
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。
注意， 题目给了提示了，要用64位的整数。
记忆化搜索方法
#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; }
![洛谷P1216-[USACO1.5][IOI1994]数字三角形 Number Triangles（DP）](https://ucc.alicdn.com/images/user-upload-01/20200726134345890.png?x-oss-process=image/resize,h_160,m_lfit)