UVAlive 4794 Sharing Chocolate 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=12055 思路:   设d[S][r][c]表示形如r*c的矩形是否可以划分为S中的子集,10表示可否. 转移方程: d[S][r][c] = d[S0][r0][c] || d[S0][r][c0]  优化:    首先注意到S r c三者知二求一,所以将状态优化为d[S][x]表示有短边x的矩形是否可以…

Sharing Chocolate Chocolate in its many forms is enjoyed by millions of people around the world every day. It is a truly universal candy available in virtually every country around the world. You find that the only thing better than eating chocolate…

n的规模可以状压,f[x][y][S]表示x行,y列,S集合的巧克力能否被切割. 预处理出每个状态S对应的面积和sum(S),对于一个合法的状态一定满足x*y=sum(S),实际上只有两个变量是独立的. 而且有x,y等效与y,x,那么这里取max(x,y). 转移的时候枚举S的非空真子集,横着切或者竖着切. 边间是到达一个合法的x,y,S,其中S中只有一个元素. 复杂度O(x*3^n) #include<bits/stdc++.h> using namespace std; ,Mxs = &l…

这道题目的DP思想挺先进的,用状态DP来表示各个子巧克力块.原本是要 dp(S,x,y),S代表状态,x,y为边长,由于y可以用面积/x表示出来,就压缩到了只有两个变量,在转移过程也是很巧妙,枚举S的子集s0,然后 s1=S-s0来代表除该子集的另一个集合,接下来分两种情况,如果这个子集是通过把 S保留x,切割y,则转移到dp(s0,x)和dp(s1,x),另一种情况是转移到dp(s0,y)和dp(s1,y).为了更加缩小状态,统一把转移方程的 x换成 min(x,sum[S]/x),sum为该…

大白书中的题感觉一般都比较难,能理解书上代码就已经很不错了 按照经验,一般数据较小的题目,都有可能是用状态压缩来解决的 题意:问一个面积为x×y的巧克力,能否切若干刀,将其切成n块面积为A1,A2,,,An块巧克力.(每次只能沿直线切一块巧克力) 设计状态: f(r, c, S) = 1表示r行c列的巧克力可以切成面积集合为S的若干块巧克力 分解问题: f(r, c, S) = 1当且仅当 横着切:存在1≤r0<r和S的子集S0,使得f(r0, c, S0) = f(r-r0,c, S-S0)…

题意 有一块\(x*y\)的巧克力,问能否恰好分成n块,每块个数如下 输入格式 n x y a1 a2 a3 ... an 首先\(x \times y 必然要等于 \sum\limits_{i=1}^{n}a_i\) 设集合状态为S,则转移方程为 \(f(x,y,S)=(f(x,c_0,S_0)\&\& f(x,y-c_0,S_1))\|(f(c_0,y,S_0)\&\& f(x-c_0,y,S_1)) \) 分别对应横着掰和竖着掰 由于 \(x \times y = \…

题目 https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2795 题意 x * y的巧克力,问能不能恰好切成n份(只能整数切),每块大小恰好ai 思路 明显,记忆化搜索. 这里参照刘书使用了sum来通过长计算宽 感想: 这种需要枚举子状态的题总是不敢枚举 代码 #include <algorithm> #includ…

 UVAlive 3983 Robotruck 题目: Robotruck   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description   Problem C - Robotruck Background This problem is about a robotic truck that distributes mail packages to sev…

A: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=83690#problem/A 题意:N*M的格子,从左上走到右下,要求在每个点的权值必须大于0,问起始的时候必须有多少能量 思路:二分答案 #include<iostream> #include<cstdio> #include<string> #include<cstring> #include<vector> #include&…

题目链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3540 题目大意: 给一块长x,宽y的巧克力,和一个数组A={a1, a2, …,an},问能否经过若干次切分后,得到面积分别为a1,a2,…an的n块巧克力.每次切分只可以选择一块巧克力,将其分为两半,如下图,3×4的巧克力经过切分后,可以得到面积分别为6,3,2,1的巧克力.…

传送门 记忆化搜索. 在下觉得sxy大佬的代码写得相当好,通篇的骚操作(因为我都不会呀),%%% 学到了 预处理每个状态的值.以前的我都是zz地枚举每一位.. for(int i=1;i<(1<<n);++i) { x=(i&(-i)); if(i==x) continue; tot[i]=tot[x]+tot[i^x]; } f[r][s]==1表示搜索过且不合法,f[r][s]==2表示搜索过且合法   if(f[r][s]) return f[r][s]-1; 若是只剩一块…

状态压缩·一 题目传送:#1044 : 状态压缩·一 AC代码: #include <map> #include <set> #include <list> #include <cmath> #include <deque> #include <queue> #include <stack> #include <bitset> #include <cctype> #include <cstdi…

Sharing Chocolate Chocolate in its many forms is enjoyed by millions of people around the world every day. It is a truly universal candy available in virtually every country around the world. You find that the only thing better than eating chocolate…

入口 UVALive - 3882 #include<cstdio> using namespace std; ; int n,m,k,f[N]; int main(){ //f[i]表示第i次要删的数字 &&n){ //1 //f[1]=0; //for(int i=2;i<=n;i++) f[i]=(f[i-1]+k)%i; //int ans=(m-k+f[n]+1)%n; //2 //int s=0; //for(int i=2;i<=n;i++) s=(s…

POJ 3342 Party at Hali-Bula / HDU 2412 Party at Hali-Bula / UVAlive 3794 Party at Hali-Bula / UVA 1220 Party at Hali-Bula(树型动态规划) Description Dear Contestant, I'm going to have a party at my villa at Hali-Bula to celebrate my retirement from BCM. I w…

UVA 1394 And Then There Was One / Gym 101415A And Then There Was One / UVAlive 3882 And Then There Was One / POJ 3517 And Then There Was One / Aizu 1275 And Then There Was One (动态规划,思维题) Description Let's play a stone removing game. Initially, n ston…

UVAlive 3135 Argus Argus Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, fin…

UVAlive 3026 Period 题目: Period   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive…

UVAlive 3942 Remember the Word 题目: Remember the Word   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description Neal is very curious about combinatorial problems, and now here comes a problem about words. Kn…

UVAlive 4329 Ping pong 题目: Ping pong Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description N(3N20000) ping pong players live along a west-east street(consider the street as a line segment). Each player ha…

UVAlive 3027 Corporative Network 题目:   Corporative Network Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 3450   Accepted: 1259 Description A very big corporation is developing its corporative network. In the beginning each of the N ent…

UVAlive X-Plosives 思路:    “如果车上存在k个简单化合物,正好包含k种元素,那么他们将组成一个易爆的混合物”  如果将(a,b)看作一条边那么题意就是不能出现环,很容易联想到Kruskal算法中并查集的判环功能(新加入的边必须属于不同的两个集合否则出现环),因此本题可以用并查集实现.模拟装车过程即可. 代码: #include<cstdio> #include<cstring> #define FOR(a,b,c) for(int a=(b);a<(c…

UVAlive 4670 Dominating Patterns 题目:   Dominating Patterns   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description The archaeologists are going to decipher a very mysterious ``language". Now, they kno…

UVa 1628 Pizza Delivery 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=51189 思路:   本体与修缮长城一题有所相似.所以解法有相似之处. 不同之处就是本体可能会产生负情况,即送餐时间晚了客户会反过来找你要钱所以需要放弃,但修缮长城只有费用,顺手修了肯定是一个不错的选择. 依旧将区间两端与位置作为状态不过要添加一维cnt表示还需要送餐的人数.类似地定义:d[i][j][cnt][p]表示…

 UVa 1380 A Scheduling Problem 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=41557 思路:   给出一个任务调度树,单向边u->v表示u必须在v之前完成,双向边u-v表示无所谓方向. 题目给出定理,首先dfs求得忽略无向边后的最长链点数k,那么问题就是判断是否可以通过无向边定向从而使得最长链点数不超过k.用dp的判断. 设f[i]表示以i为根的子树中所有的边定向后最长链点数不超过…

UVa 12170 Easy Climb 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=24844 思路:  引别人一个题解琢磨一下: from:http://blog.csdn.net/glqac/article/details/45257659 代码: #include<iostream> #include<algorithm> #define FOR(a,b,c) for(int a=(b);a…

UVa 12099  The Bookcase 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=42067 思路:   将n本书分配到三层,使得形成的书架w*h最小 提前将书籍按照高度排序,因为无论第一本书(最高的书)无论放在那一层都会被考虑到,所以规定将它放在第一层,且第二层比第三层高. 因为从大到小排序的关系,只要jk==0那么新加入的书i就是该层的高度,否则高度不变. 设d[i][j][k]表示考虑过i本书第二…

UVa 10618 Fun Game 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=36035 思路:   一圈人围坐,给出小球的传递序列,求解最少有多少个人. 问题简单化:如果有一排人,单向传递,给出序列求解最少多少人.那么问题就是:求解一个最短序列使得给出的所有序列都是该序列的连续子序列. 再分别解决以下问题: 第一个人可以向左向右传递: 每个传递序列有两种情况,分别是正序与逆序.因为不清楚当前序列是向左传还是…

UVa 10618 Fixing the Great Wall 题目:  http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=36139 思路:   数轴上有n个点需要修复,每个点有信息c,x,d 表示位于x且在t时修缮的费用是c+d*t,找一个修缮序列使n个点能全部修缮且有费用最小. 可以发现:在任意时刻,修缮完的点都是连续的,因为修缮不需要时间,将一些点“顺手”修缮了肯定不差. d[i][j][k],表示已经将i-j个点修缮…

UVa 1627 Team them up! 题目: Team them up! Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description   Your task is to divide a number of persons into two teams, in such a way, that: everyone belongs to one of…