题目传送门 /* 贪心:每次把一个丢掉,选择最小的.累加求和,重复n-1次 */ /************************************************ Author :Running_Time Created Time :2015-8-1 13:20:01 File Name :A.cpp *************************************************/ #include <cstdio> #include <algori…

吐槽:一辈子要在DIV 2混了. A,B,C都是简单题,看AC人数就知道了. A:如果我们定义数组为N*N的话就不用考虑边界了 #include<iostream> #include <string> #include <vector> #include<cstring> #include<cstdio> #include<cmath> #include<string> #include<algorithm>…

B 树形dp 组合的思想. Z队长的思路. dp[i][1]表示以i为跟结点的子树向上贡献1个的方案,dp[i][0]表示以i为跟结点的子树向上贡献0个的方案. 如果当前为叶子节点,dp[i][0] = 1,(颜色为1,可以断开与父节点的连接,颜色为0,不断开,方案恒为1),dp[i][1] = co[i](i节点的颜色). 非叶子节点:将所有孩子节点的dp[child][0]乘起来为sum,孩子贡献为0的总方案. 当前颜色为0时, dp[i][1] += sum/dp[child][0]*dp…

题意:给了一棵树以及每个节点的颜色,1代表黑,0代表白,求将这棵树拆成k棵树,使得每棵树恰好有一个黑色节点的方法数 解法:树形DP问题.定义: dp[u][0]表示以u为根的子树对父亲的贡献为0 dp[u][1]表示以u为根的子树对父亲的贡献为1 现在假设u为白色,它的子树有x,y,z,那么有 dp[u][1]+=dp[x][1]*dp[y][0]*dp[z][0]+dp[x][0]*dp[y][1]*dp[z][0]+dp[x][0]*dp[y][0]*dp[z][1] dp[u][0]+=d…

题目链接 D. Appleman and Tree time limit per test :2 seconds memory limit per test: 256 megabytes input :standard input output:standard output Appleman has a tree with n vertices. Some of the vertices (at least one) are colored black and other vertices a…

题目链接 A. Appleman and Easy Task time limit per test:2 secondsmemory limit per test:256 megabytesinput:standard inputoutput:standard output Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can…

C. Appleman and a Sheet of Paper   Appleman has a very big sheet of paper. This sheet has a form of rectangle with dimensions 1 × n. Your task is help Appleman with folding of such a sheet. Actually, you need to perform q queries. Each query will hav…

题目: C. Appleman and Toastman time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Appleman and Toastman play a game. Initially Appleman gives one group of n numbers to the Toastman, then they s…

数学家伯利亚在<怎样解题>里说过的解题步骤第二步就是迅速想到与该题有关的原型题.(积累的重要性!) 对于这道题,可以发现其实和huffman算法的思想很相似(可能出题人就是照着改编的).当然最后只是输出cost,就没必要建树什么的了.只要理解了huffman算法构造最优二叉树的思路,就按那么想就知道每个a[i]要加多少次了. 当然这道题没想到这些也可以找出规律的,就是一种贪心思想. #include<iostream> #include<cstdio> #include…

题目: B. Appleman and Card Game time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Applem…