标签:imu follow sam initial out integer term 转移 cond
There is a stone game.At the beginning of the game the player picks n
piles of stones in a line.
The goal is to merge the stones in one pile observing the following rules:
You are to determine the minimum of the total score.
Example 1:
Input: [3, 4, 3]
Output: 17
Example 2:
Input: [4, 1, 1, 4] Output: 18 Explanation: 1. Merge second and third piles => [4, 2, 4], score = 2 2. Merge the first two piles => [6, 4],score = 8 3. Merge the last two piles => [10], score = 18
思路:
区间动态规划.
设定状态: f[i][j] 表示合并原序列 [i, j] 的石子的最小分数
状态转移: f[i][j] = min{f[i][k] + f[k+1][j]} + sum[i][j], sum[i][j] 表示原序列[i, j]区间的重量和
边界: f[i][i] = 0, f[i][i+1] = sum[i][i+1]
答案: f[0][n-1]
public class Solution { /** * @param A an integer array * @return an integer */ int search(int l, int r, int[][] f, int[][] visit, int[][] sum) { if (visit[l][r] == 1) return f[l][r]; if (l == r) { visit[l][r] = 1; return f[l][r]; } f[l][r] = Integer.MAX_VALUE; for (int k = l; k < r; k++) { f[l][r] = Math.min(f[l][r], search(l, k, f, visit, sum) + search(k + 1, r, f, visit, sum) + sum[l][r]); } visit[l][r] = 1; return f[l][r]; } public int stoneGame(int[] A) { if (A == null || A.length == 0) { return 0; } int n = A.length; // initialize int[][] f = new int[n][n]; int[][] visit = new int[n][n]; for (int i = 0; i < n; i++) { f[i][i] = 0; } // preparation int[][] sum = new int[n][n]; for (int i = 0; i < n; i++) { sum[i][i] = A[i]; for (int j = i + 1; j < n; j++) { sum[i][j] = sum[i][j - 1] + A[j]; } } return search(0, n-1, f, visit, sum); } }
标签:imu follow sam initial out integer term 转移 cond
原文地址:https://www.cnblogs.com/FLAGyuri/p/12078350.html