题目链接: D. Kindergarten time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In a kindergarten, the children are being divided into groups. The teacher put the children in a line and associated e…

[Codeforces 1201D]Treasure Hunting(DP) 题面 有一个n*m的方格,方格上有k个宝藏,一个人从(1,1)出发,可以向左或者向右走,但不能向下走.给出q个列,在这些列上可以向上走,其他列不能向上走.可以重复经过同一个点.求从(1,1)出发,经过所有宝藏的最短路径长度 \(n,m,k,q \leq 2 \times 10^5\) 分析 贪心考虑,我们应该按照行一层一层的走.每一行应该从最左的宝藏走到最右的宝藏,或者从最右的宝藏走到最左的宝藏,然后找最近的一个可以向…

题意:给n个数,分成若干个连续组,每组获益为max-min,输出最大获益. 参考:http://blog.csdn.net/keshuai19940722/article/details/40873581 参考的链接里说得很明白了,我的dp[i][0]是升序,dp[i][1]是降序,习惯而已. 这题关键点就是,如果a[i-1]<a[i]>a[i+1],显然3个分开(a[i]归左或右)不比在一起差. #include <cstdio> #include <cstring>…

D. Kindergarten Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/484/D Description In a kindergarten, the children are being divided into groups. The teacher put the children in a line and associated each child wi…

D. Kindergarten     In a kindergarten, the children are being divided into groups. The teacher put the children in a line and associated each child with his or her integer charisma value. Each child should go to exactly one group. Each group should b…

贪心观察+DP决策. 首先需要观察到一个结论:分割后的每一段肯定是单调增或者单调减的. 然后可以根据dp来决策如何分割价值最多. dp[i][0]表示放完第i个,最后一段是递减的情况下的最大价值 dp[i][1]表示放完第i个,最后一段是递增的情况下的最大价值 #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; ; ]; lon…

大意: 给定序列, 求划分为若干段, 使得总贡献最大, 每段的贡献为max-min 可以发现最优解一定是连续一段递增或递减, 然后dp即可. #include <iostream> #include <cstdio> #define REP(i,a,n) for(int i=a;i<=n;++i) using namespace std; typedef long long ll; const int N = 1e6+10; int n, a[N]; ll dp[2][N];…

http://www.codeforces.com/gym/100827/attachments Hill Number Time Limits:  5000 MS   Memory Limits:  200000 KB 64-bit interger IO format:  %lld   Java class name:  Main Description A Hill Number is a number whose digits possibly rise and then possibl…

题目链接:http://codeforces.com/contest/710/problem/E 题意:开始文本为空,可以选择话费时间x输入或删除一个字符,也可以选择复制并粘贴一串字符(即长度变为两倍),问要获得长度为n的串所需最少的时间. 思路:dp[i]表示获得长度为i的串所需要的最短时间,分i为奇数和偶数讨论. #include<bits/stdc++.h> using namespace std; const int N=1e7+3; typedef long long ll; ll…

题目链接:http://codeforces.com/contest/699/problem/C dp[i][j]表示第i天做事情j所得到最小的假期,j=0,1,2. #include<bits/stdc++.h> using namespace std; const int INF=0x3f3f3f3f; int dp[105][3]; int main() { int n; scanf("%d",&n); memset(dp,INF,sizeof(dp)); d…