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<…

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…

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"…

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…

密码学课上老师介绍了这样一个问题,囚徒问题(100 prisoners problem):一百个囚徒被关在牢房里,典狱长给他们最后一次机会,100人依次进入一个有100个抽屉的牢房,每个抽屉置乱放入1~100的号码,每个人依次打开50个抽屉,如果每个人都能找到自己的号码(0~100),则所有人被释放,反之所有人都会被杀死. 典型的算法给出的被释放概率几乎为零,但是有这样一种方法:第i号囚徒打开第i号抽屉,如果这个抽屉放的是自己的号码牌,ok pass,如果不是,则打开该抽屉中号码牌对应的抽屉,依…

Gym 101047E Escape from Ayutthaya Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u  Practice  Description standard input/output Ayutthaya was one of the first kingdoms in Thailand, spanning since its foundation in 1350 to…