【Title description】
N piles of stones are arranged in a line. Now combine the stones into a pile in an orderly manner. It is stipulated that only 2 adjacent piles of stones can be selected to merge into a new pile at a time, and the number of new piles of stones is recorded as the cost of the merger. Calculate the minimum cost of merging N piles of stones into one pile.
For example: 1 2 3 4, there are many ways to combine
1 2 3 4 => 3 3 4(3) => 6 4(9) => 10(19)
1 2 3 4 => 1 5 4(5) => 1 9(14) => 10(24)
1 2 3 4 => 1 2 7(7) => 3 7(10) => 10(20)
The total cost in parentheses shows that the first method has the lowest cost. Now, given the number of n piles of stones, calculate the minimum combined cost.
Input
Line 1: N (2 <= N <= 100)
Line 2 - N + 1: Number of N piles of stones (1 <= A i <= 10000)
Line 2 - N + 1: Number of N piles of stones (1 <= A i <= 10000)
Output
output minimum merge cost
Sample Input
4 1 2 3 4Sample Output
19
【analyze】
Interval DP template questions.
Preprocess the interval sum, then enumerate the interval and breakpoint of each length, and transfer.
f[i][j] = min(f[i][j], f[i][k]+f[k+1][j]+sum[j]-sum[i-1])
【Code】
#include <iostream> #include <cstdlib> #include <cstdio> #include <cstdio> #include <cstring> using namespace std; typedef long long LL; const int maxn = 1000 + 100; #define INF 0x3f3f3f3f intmain () { int n; scanf("%d", &n); int sum[maxn]; int f[maxn][maxn]; memset(f, 0, sizeof(f)); for (int i = 1; i <= n; i++) { int x; scanf("%d", &x); sum[i] = sum[i-1]+x; } for ( int len = 2 ; len <= n; len ++ ) { for (int i = 1; i <= n; i++) { int j = i + len- 1 ; f[i][j] = INF; for (int k = i; k < j; k++) f[i][j] = min(f[i][j], f[i][k]+f[k+1][j] + sum[j] - sum[i-1]); } } printf("%d\n", f[1][n]); }