1389 算个简单的树形DP吧 不知道是不是数据太水 竟然一A了 就是对于当前节点有没有被选中就行选最优 有没有被选中的意思是有没有与它相连的边被选中 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cmath> using name…

Labyrinth Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1507    Accepted Submission(s): 520 Problem Description 度度熊是一只喜欢探险的熊,一次偶然落进了一个m*n矩阵的迷宫,该迷宫只能从矩阵左上角第一个方格开始走,只有走到右上角的第一个格子才算走出迷宫,每一次只能走一格,…

Break Standard Weight Time Limit: 2 Seconds                                     Memory Limit: 65536 KB                             The balance was the first mass measuring instrument invented. In its traditional form, it consists of a pivoted horizon…

LightOJ 1033  Generating Palindromes(dp) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/A 题目: Description By definition palindrome is a string which is not changed when reversed. "MADAM" is a nice example of palindrome.…

lightOJ 1047   Neighbor House (DP) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/C 题目: Description The people of Mohammadpur have decided to paint each of their houses red, green, or blue. They've also decided that no two n…

C. Boredom time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a ga…

题目链接:http://poj.org/problem?id=2479 题意:求所给数列中元素值和最大的两段子数列之和. 分析:从左往右扫一遍,b[i]表示前i个数的最大子数列之和. 从右往左扫一遍,c[i]表示后i个数的最大子数列之和. ans=max(ans,b[i]+c[i+1])0<i<n dp方程为 sum=max(sum+a[i],a[i]) dp[i]=max(dp[i-1],sum) #include <cstdio> #include <cstring>…

UVA11125 - Arrange Some Marbles(dp) option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2066" target="_blank" style="">题目链接 题目大意:给你n种不同颜色的弹珠.然后给出每种颜色的弹珠的个数,如今要求你将这些弹珠排序,要求同样颜色的部分最多3个.然后同样颜色的弹珠称…

题意:给出按照x坐标排序的n个点,让我们求出从最左端点到最右短点然后再回来,并且经过所有点且只经过一次的最短路径. 分析:这个题目刘汝佳的算法书上也有详解(就在基础dp那一段),具体思路如下:按照题目的描述比较难考虑,不如把这个问题想成两个人,分别从最左端走到最右端并且不走到重复的点所需要的最小路程,我们定义dp[i,j]表示第一个人走到第i个位置,第二个人走到第j个位置所耗费的路程,并且让第一个人的位置大于第二个人的位置,且前i个城市都已经被走过,那这样的话能走的点只有i+1,i+2,……n这…

有形如下图所示的数塔,从顶部出发,在每一结点可以选择向左走或是向右走,一起走到底层,要求找出一条路径,使路径上的值最大. 样例输入: 5 13 11 8 12 7 26 6 14 15 8 12 7 13 24 11 样例输出: 86(13->8->26->15->24) #include <iostream> #include <cstdio> #include <cstring> #define maxn 105 using namespac…