题目传送门
解题思路:
f[i][j]表示到(i,j)的方案数,则f[i][j] = f[i-1][j] + f[i][j-1];
代码:
1 #include<cstdio> 2 #include<iostream> 3 4 using namespace std; 5 6 long long n,m,n1,m1,f[21][21]; 7 bool a[101][101]; 8 9 int main() { 10 scanf("%lld%lld%lld%lld",&n,&m,&n1,&m1); 11 a[n1][m1] = a[n1-2][m1-1] = a[n1-1][m1-2] = a[n1-2][m1+1] = a[n1-1][m1+2] = a[n1+1][m1-2] = a[n1+2][m1-1] = a[n1+1][m1+2] = a[n1+2][m1+1] = 1; 12 f[0][0] = 1; 13 for(int i = 1;i <= m; i++) 14 if(a[0][i]) 15 break; 16 else 17 f[0][i] = f[0][i-1]; 18 for(int i = 1;i <= n; i++) 19 if(a[i][0]) 20 break; 21 else 22 f[i][0] = f[i-1][0]; 23 for(int i = 1;i <= n; i++) 24 for(int j = 1;j <= m; j++) 25 if(!a[i][j]) 26 f[i][j] = f[i-1][j] + f[i][j-1]; 27 printf("%lld",f[n][m]); 28 return 0; 29 }
看了题解后,发现可以优化空间,状态再压一维.
f[j] = f[j] + f[j-1]; (j表示列)
咋来的呢? 再每一个f[j]更新前,其实原来储存的是上一行的第j列的答案.举个例子:我已经枚举到第3行了,再f[j]没更新前,f[j]里储存的其实是第2行第j列的答案.f[j-1]跟朴素一样.
代码:
1 #include<cstdio> 2 #include<iostream> 3 4 using namespace std; 5 6 long long n,m,n1,m1,f[22]; 7 bool a[101][101]; 8 9 int main() { 10 scanf("%lld%lld%lld%lld",&n,&m,&n1,&m1); 11 n++;m++;n1++;m1++;//压维后如果不加1,运算期间下标会成负数 12 a[n1][m1] = a[n1-2][m1-1] = a[n1-1][m1-2] = a[n1-2][m1+1] = a[n1-1][m1+2] = a[n1+1][m1-2] = a[n1+2][m1-1] = a[n1+1][m1+2] = a[n1+2][m1+1] = 1; 13 f[1] = 1; 14 for(int i = 1;i <= n; i++) 15 for(int j = 1;j <= m; j++) 16 if(!a[i][j]) 17 f[j] += f[j-1]; 18 else 19 f[j] = 0; 20 printf("%lld",f[m]); 21 return 0; 22 }
//好像并不优化时间