The merchant Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 6864   Accepted: 2375 Description There are N cities in a country, and there is one and only one simple path between each pair of cities. A merchant has chosen some paths and w…

The merchant Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Problem Description There are N cities in a country, and there is one and only one simple path between…

http://poj.org/problem?id=1651 题意:给出n个数字,每取中间一个数,就会使得权值加上中间这个数和两边的乘积,求取剩两个数最少的权值是多少. 思路:区间dp. 一开始想了挺久还是写不出方程,做了点别的事回来再想就突然觉得很简单了. 一开始使得长度为1和2的区间dp[i][j]为0. 然后dp[i][j] = min(dp[i][k] + dp[k][j] + w[k] * w[i] * w[j]). 枚举的k为中间拿掉的数,然后和左右的区间和相加就是最后答案了. #i…

You are given two integers l l and r r (l≤r l≤r ). Your task is to calculate the sum of numbers from l l to r r (including l l and r r ) such that each number contains at most k k different digits, and print this sum modulo 998244353 998244353 . For…

http://poj.org/problem?id=2796 题意:给出n个数,问一个区间里面最小的元素*这个区间元素的和的最大值是多少. 思路:只想到了O(n^2)的做法. 参考了http://www.cnblogs.com/ziyi--caolu/archive/2013/06/23/3151556.html的写法,用单调栈可以优化到O(n). 对于每个元素,维护这个元素向前延伸比它大的有多少个,向后延伸比它小的有多少个.即该元素是处于山谷. 那么如何用单调栈维护这个呢? 首先这个栈是单调递…

http://poj.org/problem?id=3318 题意:问A和B两个矩阵相乘能否等于C. 思路:题目明确说出(n^3)的算法不能过,但是通过各种常数优化还是能过的. 这里的随机算法指的是随机枚举矩阵C的一个位置,然后通过A*B计算是否能够得到矩阵C相应位置的数,如果不等,就直接退出了,如果跑过一定的数量后能够相等,那么就可以判断这个矩阵C等于A*B的.第一次见这样的题目...有点新奇. 暴力算法: #include <cstdio> using namespace std; ][]…

"Holiday is coming, holiday is coming, hurray hurray!" shouts Joke in the last day of his college. On this holiday, Joke plans to go to his grandmother's house located in Schematics village. Joke's grandmother's house is more than a hundred year…

http://www.lydsy.com/JudgeOnline/problem.php?id=1026 1026: [SCOI2009]windy数 Time Limit: 1 Sec  Memory Limit: 162 MBSubmit: 5561  Solved: 2493[Submit][Status][Discuss] Description windy定义了一种windy数.不含前导零且相邻两个数字之差至少为2的正整数被称为windy数. windy想知道,在A和B之间,包括A和B…

Crazy Search Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 32483   Accepted: 8947 Description Many people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a gi…

题目链接:http://poj.org/problem?id=2533 思路分析:该问题为经典的最长递增子序列问题,使用动态规划就可以解决: 1)状态定义:假设序列为A[0, 1, .., n],则定义状态dp[i]为以在所有的递增子序列中以A[i]为递增子序列的最后一个数字的所有递增子序列中的最大长度: 如:根据题目,在所有的以3结尾的递增子序列有[3]和[1, 3],所以dp[2] =2; 2)状态转移方程:因为当A[j] < A[i]时(0<= j < i),dp[i] = Max…