题目链接:http://hihocoder.com/problemset/problem/1636 题目大意:有n堆石头,每次只能合并l~r堆,每次合并的花费是要合并的石子的重量,问你合并n堆石子的最小花费,若不能合并则输出0. 解题思路: 这算是石子合并的加强版了吧,原来石子合并是只能两堆两堆地合并,而现在对合并堆数加了一个范围[l,r].这题我看到的时候也没什么想法,看了题解才会的,而且也看的不是特别懂. 首先定义数组dp[i][j][k]表示将[i,j]的石子合并成k堆需要的最小花费. 那…

思路: 区间dp.dp[l][r][k]表示把区间[l, r]的石子合并成k堆所需要的最小代价. 实现: #include <iostream> #include <cstring> using namespace std; const int INF = 0x3f3f3f3f; ; int a[N], sum[N], dp[N][N][N]; int n, L, R; int dfs(int l, int r, int k) { ? : INF; == k) ; ) return…

#1636 : Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the begi…

#1636 : Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the begi…

hihoCoder 1636 Pangu and Stones 思路:区间dp. 状态:dp[i][j][k]表示i到j区间合并成k堆石子所需的最小花费. 初始状态:dp[i][j][j-i+1]=0 状态转移: 如果k等于1,dp[i][j][1]=min(dp[i][j][1],dp[i][k][s-1]+dp[k+1][j][1]+sum[j]-sum[i-1])(i<=k<j) s从l到r,因为合并成一堆的时候是有堆数限制的 如果k大于1,dp[i][j][k]=min(dp[i][j…

#1636 : Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the begi…

Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the beginning, t…

Pangu and Stones HihoCoder - 1636 题意 给你\(n\)堆石子,每次只能合成\(x\)堆石子\((x\in[L, R])\),问把所有石子合成一堆的最小花费. 思路 和合石子的那题很像,多加了的一个限制,所有我们可以想到要多开一维数组来计算. \(dp[i][j][x]:\)表示区间\([i, j]\)的范围内有\(x\)堆石子. 然后我们要分成两类讨论(\(sum[i]\)表示前\(i\)堆石子的和) \(1\).\(dp[i][j][1] = min(dp[i…

#1636 : Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the begi…

题意:N堆石子,每次可以合并连续的长度从L到R的若干堆石子为1堆,费用为选择的石子总个数,求将N堆合并成1堆的最小总花费,无解输出0 思路:dp[i][j][k]表示将i到j这段区间合并为k堆的最小代价 \[ 初始条件 dp[i][j][j-i+1]=0 \] \[ dp[i][j][k]=min(dp[i][x][y-1]+dp[x+1][j][1]+s[j]-s[i-1] (k=1,i<=x<=j-1,L<=y<=R) \] \[ dp[i][j][k]=min(dp[i][x…