A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 989    Accepted Submission(s): 371 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie…

题目链接:hdu 3183 A Magic Lamp 题目大意:给定一个字符串,然后最多删除K个.使得剩下的组成的数值最小. 解题思路:问题等价与取N-M个数.每次取的时候保证后面能取的个数足够,而且取的数最小,查询最小的操作用RMQ优化. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 10005; int N, M, d[m…

hdu 3183 A Magic Lamp RMQ ST 坐标最小值 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3183 题目大意: 从给定的串中挑出来m个数使得剩余的数字最小,串的序列不能改变 思路: 将问题转化为求在n个数中挑选n-m个数,使之最小.假设最极端的情况,所有最大的数字都在左侧,占据了m个位置,那么我们需要挑选的最小的数字的第一位就是在m+1位上.依次类推,第二位在m+2位上,最后一位也就在原串的最后一位上.反过来,假设最大的数…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3183 Problem DescriptionKiki likes traveling. One day she finds a magic lamp, unfortunately the genie in the lamp is not so kind. Kiki must answer a question, and then the genie will realize one of her d…

A Magic Lamp                                                                               Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Kiki likes traveling. One day she finds a magic lamp, u…

A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie in the lamp is not so kind. Kiki must answer a question…

原题目 A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3964    Accepted Submission(s): 1605 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the…

RMQ. /* 3183 */ #include <cstdio> #include <cstring> #include <cstdlib> #define MAXN 1005 char s[MAXN], ans[MAXN]; int dp[MAXN][MAXN]; int n,len,m; int min(int x, int y) { return s[x]<=s[y] ? x:y; } void RMQ_init() { int i, j, k; ; i&…

直接模拟   如果后一位比前一位小,那就一直 向前 pop()掉 维护他单调递增: #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<stdio.h> #include<vector> using namespace std; ]; ],num; int main( ) { while( scanf("%s%d&…

第一种做法是贪心做法,只要前面的数比后面的大就把他删掉,这种做法是正确的,也比较好理解,这里就不说了,我比较想说一下ST算法,RMQ的应用 主要是返回数组的下标,RMQ要改成<=(这里是个坑点,取连续数是可以的),他的转移方程为x = dp[i-1][j],y = dp[i-1][j+1<<(i-1)]; dp[i][j] = a[x] <= a[y] ? x : y; 这里既是转化为取n-m个数,首先在1 ~ m+1中必然选一个数,所以这个数就是n-m数中第一个数,记录所选的位置…