Problem In a kingdom there are prison cells (numbered 1 to P) built to form a straight line segment. Cells number i and i+1 are adjacent, and prisoners in adjacent cells are called "neighbours." A wall with a window separates adjacent cells, and…

GCJ1C09C - Bribe the Prisoners Problem In a kingdom there are prison cells (numbered 1 to P) built to form a straight line segment. Cells number i and i+1 are adjacent, and prisoners in adjacent cells are called "neighbours." A wall with a windo…

题目链接: http://www.spoj.com/problems/GCJ1C09C/ 题意: In a kingdom there are prison cells (numbered 1 to P) built to form a straight line segment. Cells number i and i+1 are adjacent, and prisoners in adjacent cells are called "neighbours." A wall wi…

Problem In a kingdom there are prison cells (numbered 1 to P) built to form a straight line segment. Cells number i and i+1 are adjacent, and prisoners in adjacent cells are called "neighbours." A wall with a window separates adjacent cells, and…

一个监狱里有P个并排着的牢房,从左往右一次编号为1,2,-,P.最初所有牢房里面都住着一个囚犯.现在要释放一些囚犯.如果释放某个牢房里的囚犯,必须要贿赂两边所有的囚犯一个金币,直到监狱的两端或者空牢房为止.现在要释放a1,a2,...,aQ号囚犯,如何选择释放的顺序,使得使用的金币最少. 思路: 其中很重要的一点:释放了某个囚犯以后,就把连续的牢房分成了没有任何关系的两段. 只要枚举出所有的释放囚犯的顺序即可,复杂度为 O(Q3). 利用动态规划枚举所有的情况的时候,我们有2种方法: 方法1.(…

区间DP.dp[i][j]表示第i到第j个全部释放最小费用. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; const int INF=0x7FFFFFFF; +; int T,P,Q; int a[maxn]; int dp[maxn][maxn]; void read() { scanf("%d%d"…

看来我还是太菜了,这么一道破题做了那么长时间...... 传送门 分析 我首先想到的是用状压dp来转移每一个人是否放走的状态,但是发现复杂度远远不够.于是我们考虑区间dp,dpij表示i到j区间的所有罪犯全部放走的最小花费,于是我们可以将一个区间(i,j)分为(i,k-1),(k+1,j)和k这个点,表示先取走点k的人,这样这个区间就被分成了两个,于是我们便可以转移的.详见代码. 代码 #include<iostream> #include<cstdio> #include<…

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2588: Spoj 10628. Count on a tree Time Limit: 12 Sec  Memory Limit: 128 MBSubmit: 5217  Solved: 1233[Submit][Status][Discuss] Description 给定一棵N个节点的树,每个点有一个权值,对于M个询问(u,v,k),你需要回答u xor lastans和v这两个节点间第K小的点权.其中lastans是上一个询问的答案,初始为0,即第一个询问的u是明文. Input 第一…

题目 Source http://www.spoj.com/problems/DQUERY/en/ Description Given a sequence of n numbers a1, a2, ..., an and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elem…