ZOJ 3605 大体意思就是 找出随机选了K个交换后 石子在第i个罐子里的概率最大 也就是可能的总数最大 这样就可以写出递推方程 dp[i][j][k] += dp[i-1][e][k]; (0<e<j&& k!=a[j]&&k!=b[j]) dp[i][j][a[j]]+=dp[i-1][e][b[j]] dp[i][j][b[j]]+=dp[i-1][e][a[j]]; dp[i][j][k]表示选第i次交换时 选的为第j个 而且石子在第K个罐子里的可能…

ZOJ Problem Set - 3822Domination(DP) problemCode=3822">题目链接 题目大意: 给你一个n * m的棋盘,每天都在棋盘上面放一颗棋子.直到这个棋盘上的每行每列都有至少有一颗棋子.求要用的天数的期望. 解题思路:         先求出不同摆法的棋盘的概率,然后在和天数相乘就是期望.         我们将棋盘划分为四个部分:当中一部分为每行没列都至少有一个棋子.         然后得出状态转移方程:         dp[x][y][k…

题意:切一个凸边行,如果不是凸包直接输出.然后输出最小代价的切割费用,把凸包都切割成三角形. 先判断是否是凸包,然后用三角形优化. dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+w[i][k]+w[j][k]); w[i][j]代表i到j点的切割费用. dp[i][j]:表示以i到j点的最小费用.则可把凸边行分成三个部分的费用.两个凸边行(i,k),(k,j)和两条边的费用(i,k),(j,k),k为枚举的三角形顶点. Zoj 3537 Cake (DP_最优三…

B - Battle Ships Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3623 Appoint description:   Description Battle Ships is a new game which is similar to Star Craft. In this game, the enemy builds a…

Breaking Strings Time Limit: 2 Seconds        Memory Limit: 65536 KB A certain string-processing language allows the programmer to break a string into two pieces. Since this involves copying the old string, it costs n units of time to break a string…

ZOJ - 3777 就是一个入门状压dp期望 dp[i][j] 当前状态为i,分数为j时的情况数然后看代码 有注释 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #inclu…

转自:http://blog.csdn.net/a497406594/article/details/38442893 Kill the Monsters Time Limit: 7 Seconds Memory Limit: 32768 KB In order to celebrate the 8th anniversary of ZOJ, watashi introduces a strange game to other ZJU ACM team members. The board of…

Painting Storages Time Limit: 2 Seconds      Memory Limit: 65536 KB There is a straight highway with N storages alongside it labeled by 1,2,3,...,N. Bob asks you to paint all storages with two colors: red and blue. Each storage will be painted with e…

ZOJ Problem Set - 3822 Domination Time Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a lar…

题意: 给你三个均匀k面筛子. 分别有k1 k2 k3个面,每个面朝上的概率是相等的. 如果第一个筛子出现a第二个筛子出现b第三个筛子出现c那么置零. 否则在当前和加上三个点数之和. 求当前和大于n需要的步数的期望. 思路: 一开始状态转移搞错了,手推公式交了WA,后来想了想状态转移的过程发现每个状态都跟0状态有关系,但是dp[0]不确定,但是幸运的是这是一个线性变换,所以状态转移的时候记录一下dp[0]的系数,最后移项输出就好了. dp[i]=dp[i+x]*(k1*k2*k3);(x=i+j…