https://acm.sdut.edu.cn/onlinejudge2/index.php/Home/Contest/contestproblem/cid/3016/pid/1729
Stone merger
Time Limit: 1000 ms Memory Limit: 65536 KiB
Problem Description
A circle around the playground placed stones piled n. Stone is to have the order to merge into a pile. Provisions can only choose two adjacent pile into a new pile of stones, a pile of stones and a few new record for the combined score. Try to design an algorithm to calculate the n combined into a pile of stones piled minimum score and maximum score.
For a given n gravel pile, merge into a pile calculating the minimum and maximum score score.
For a given n gravel pile, merge into a pile calculating the minimum and maximum score score.
Input
First data input line is a positive integer n, 1≤n≤100, n represents stones piled there. The second row has a number of n, respectively, represent the number of each pile of stones.
Output
There are two lines of output data, the number of line 1 is the minimum score, the second number of rows is the maximum score.
Sample Input
4 4 4 5 9
Sample Output
43 54
Algorithm finished the exam (the title a bit simple luck AK ..), but the problem was bothering me for a long time, the online answer is very vague, so it is here to record it.
This question difficulty is: if the ring into a linear (straight-line there are many answers, not elaborated here).
The core idea is this: by the number of 2n-1 to become converted into a linear problem .
Was puzzled to see these lines for a long time, has been very understanding, but after drawing I understand.
1. 我们为了将环形变为直线，必须规定转动顺序，这里采用逆时针转动，且以 i 作为起点，j作为终点。(右下图）
2. 当规定好终点了，那么这环形有4种情况，我们求在这四种情况下最下的。(右上图）
3. 关于转换成直线，比如存在 a(0) -> b(1) -> c(2) -> d(3) 与 d(3) -> a(4) -> b(5) -> c(6)。
第一条是 i=0,j=3的数组，第二条是 i=3,j=6 的数组。 这样，我们就不用返回去计算了( 3->0->1->2 ???)。
4. 四种情况分别是数组中的四行，每行最后一个代表遍历的结果，我们最后只需要遍历这四个值，找到最小值即可。
源代码：
1 #include <iostream> 2 #include <stdio.h> 3 #include <algorithm> 4 #define INF 0x3f3f3f3f 5 using namespace std; 6 int Arr[300],Sum[300]; 7 int Min[300][300], Max[300][300]; 8 int main() { 9 int n; 10 cin >> n; 11 12 // 初始化数组 13 for (int i = 1; i <= n; i++) { 14 cin >> Arr[i]; 15 Arr[i + n] = Arr[i]; 16 } 17 18 // 计算最大和 19 for (int i = 1; i <= 2 * n; i++) { 20 Sum[i] = Sum[i - 1] + Arr[i]; 21 } 22 23 // 开始递归循环 24 for (int i = 2 * n-1; i >= 1; i--) { 25 for (int j = i + 1; j < i + n; j++) { 26 Min[i][j] = INF; 27 for (int k = i; k < j; k++) { 28 Min[i][j] = min(Min[i][j], Min[i][k] + Min[k + 1][j] + Sum[j] - Sum[i - 1]); 29 Max[i][j] = max(Max[i][j], Max[i][k] + Max[k + 1][j] + Sum[j] - Sum[i - 1]); 30 } 31 } 32 } 33 34 // 遍历找到最大与最小值 35 int MaxValue = 0, MinValue = INF; 36 for (int i = 1; i <= n; i++) { 37 MaxValue = max(MaxValue, Max[i][i + n - 1]); 38 MinValue = min(MinValue, Min[i][i + n - 1]); 39 } 40 41 cout << MinValue << endl << MaxValue << endl; 42 43 }