随笔分类 - 动态规划 - 区间dp
摘要：思路一：（区间dp） dp[i][j] 表示合并区间 i ~ j 所能获得的最大收益 class Solution { public: int a[505]; int dp[505][505]; int stoneGameV(vector<int>& stoneValue) { int len =
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摘要：题意 ： 给你 n 个数字，相邻的数字如果相同，则代表他们是一个块的，每次操作可以将一个块的数字变成任意一种数字，求最小操作次数，将整个区间的所有数字变成相同的 思路分析 ： 定义 dp[i][j][k] k = 0 / 1 ， dp[i][j][0] 表示将区间变成和左边数字一样的最小操作次数，
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摘要：There is one last gate between the hero and the dragon. But opening the gate isn't an easy task. There were n buttons list in a straight line in front
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摘要：When we are focusing on solving problems, we usually prefer to stay in front of computers rather than go out for lunch. At this time, we may call for
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摘要：Long long ago, there lived two rabbits Tom and Jerry in the forest. On a sunny afternoon, they planned to play a game with some stones. There were n s
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摘要：Here is a famous story in Chinese history. That was about 2300 years ago. General Tian Ji was a high official in the country Qi. He likes to play hors
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摘要：Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a regular sequence, then (S) and [
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摘要：There is a straight highway with villages alongside the highway. The highway is represented as an integer axis, and the position of each village is id
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摘要：以一个经典题目引入到正题 ： 有 n 堆石子 ， 每两堆石子合并会花费一定的价值，所花费的价值即为 两堆石子上的价值和 ， 问合并所有的石子后的最小花费 ？ 思路分析 ： 因为题干可以看成是对每个区间的 的操作，找到问题的子问题 ， 即只有一堆石子， 即区间长度为 1 的情况 ，它并不要合并 ， 所
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