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…

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…

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…

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<…

POJ 1852 Ants POJ 2386 Lake Counting POJ 1979 Red and Black AOJ 0118 Property Distribution AOJ 0333 Ball POJ 3009 Curling 2.0 AOJ 0558 Cheese POJ 3669 Meteor Shower AOJ 0121 Seven Puzzle POJ 2718 Smallest Difference POJ 3187 Backward Digit Sums POJ 3…

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…