hdu 3944 dp?】的更多相关文章

DP? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 128000/128000 K (Java/Others)Total Submission(s): 1804    Accepted Submission(s): 595 Problem Description Figure 1 shows the Yang Hui Triangle. We number the row from top to bottom 0,1,2,…a…
DP? Problem Description Figure 1 shows the Yang Hui Triangle. We number the row from top to bottom 0,1,2,…and the column from left to right 0,1,2,….If using C(n,k) represents the number of row n, column k. The Yang Hui Triangle has a regular pattern…
DP? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 128000/128000 K (Java/Others)Total Submission(s): 3126    Accepted Submission(s): 978 Problem Description Input Input to the problem will consists of series of up to 100000 data sets. For e…
题意:在杨辉三角中让你从最上面到 第 n 行,第 m 列所经过的元素之和最小,只能斜向下或者直向下走. 析:很容易知道,如果 m 在n的左半部分,那么就先从 (n, m)向左,再直着向上,如果是在右半部分,那么就是先向左斜着走再向上.这样对应到, 然后化简得到的答案就是C(n+1, m) + n - m,和C(n+1, m+1) + m.然后用Lucsa 定理就好. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000"…
Man Down Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2030    Accepted Submission(s): 743 Problem Description The Game “Man Down 100 floors” is an famous and interesting game.You can enjoy t…
给出点集,和不大于L长的绳子,问能包裹住的最多点数. 考虑每个点都作为左下角的起点跑一遍极角序求凸包,求的过程中用DP记录当前以j为当前末端为结束的的最小长度,其中一维作为背包的是凸包内侧点的数量.也就是 dp[j][k]代表当前链末端为j,其内部点包括边界数量为k的最小长度.这样最后得到的一定是最优的凸包. 然后就是要注意要dp[j][k]的值不能超过L,每跑一次凸包,求个最大的点数量就好了. 和DP结合的计算几何题,主要考虑DP怎么搞 /** @Date : 2017-09-27 17:27…
B - Monkey and Banana Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1069 Appoint description: Description A group of researchers are designing an experiment to test the IQ of a monkey. They wi…
G - FatMouse's Speed Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1160 Appoint description: Description FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4826 思路:dp[x][y][d]表示从方向到达点(x,y)所能得到的最大值,然后就是记忆化了. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define REP(i, a, b) for (int i = (a); i < (b); ++i)…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2861 题目大意:n个位置,m个人,分成k段,统计分法.S(n)=∑nk=0CknFibonacci(k) 解题思路: 感觉是无聊YY出的DP,数据目测都卡了几W组.如果不一次打完,那么直接T.$DP[i][j][k][0|1]$ 用$DP[i][j][k][0|1]$表示,$i$位置,已经安排了$j$个人,有$k$段,且$i$位置不放人/放人. 边界 $DP[0][0][0][0]=DP[0][0]…