西雅图 Day3
February 11, 2019
The Nature Conservancy Zuch Ferdana @CoastResilience Geospatial Information Officer(GIO)
Posts
西雅图 Day1
February 9, 2019
The server’s previous technical issues have caused the loss of images and data.
Intro to Neural Network
February 7, 2019
The server’s previous technical issues have caused the loss of images and data.
济南Day3 坐标型动态规划及背包
January 29, 2019
花店橱窗布置 思路 f[i][j]f[i][j]表示前i个花瓶前j个花束的最大美学价值 f[i][j]=max(f[i−1][k],f[i][j])f[i][j]=max(f[i−1][k],f[i][j]) 当然还有另外一种思路(*太强了！！！*)：
高精度模板
January 29, 2019
define N 1e5 struct bign { int len; int v[N]; // 赋值 bign=bign bign operator = (char* s) { len=strlen(s); memset(v,0,sizeof(v)); for(int i=0;i<len;i++) v[i]=s[len-i-1]-'0'; return *this; } //赋值 bign=int bign operator = (int x) { char s[N]; sprintf(s,"%d",x); return *this=s; } // 高精加 bign operator + (const bign &b) const { bign c; memset(c.
树形DP
January 29, 2019
二叉苹果树 思路 $dp[u][j表示节点u留下j条边的最大价值，每一次决策只有三种情况：剪左子树，剪右子树，两个都不剪 剪左边：dp[u][j]=dp[rson][j−1]+v[rson]dp[u][j]=dp[rson][j−1]+v[rson]，同理，剪右边：dp[u][j]=dp[lson][j−1]+v[lson]dp[u][j]=dp[lson][j−1]+v[lson] 两边都不剪：dp[u][j]=dp[lson][j]+dp[rson][k−j−2]dp[u][j]=dp[lson][j]+dp[rson][k−j−2] 代码：记忆化搜索 int f[N][N]; bool t[N][N]; int dp(int u,int k) { if(t[u][k]) return f[u][k]; t[u][k]=1; if(!
济南Day1 随记
January 28, 2019
The server’s previous technical issues have caused the loss of images and data.
济南Day2 随记
January 23, 2019
The server’s previous technical issues have caused the loss of images and data.