Base Time Limit: 1 second Space Limit: 131072 KB Score: 20Difficulty
: Level 3 Algorithm Questions
![](http://file.51nod.com/images/icon/star.png)
![](http://file.51nod.com/images/icon/plus.png)
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 each time, and the number of new piles of stones shall be 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
第1行:N(2 <= N <= 100) 2nd - N + 1: Number of N piles of stones (1 <= A[i] <= 10000)
Output
output minimum merge cost
Input example
4 1 2 3 4
Output example
19
[Code]:
#include<cstdio> #include<cstring> #include<queue> #include<iostream> #include<stack> #define maxn 105 #define maxm 50005 #define INF 0x3f3f3f3f #define ll long long using namespace std; int n; int a[maxn]; int dp[maxn][maxn]; int sum[maxn]; intmain () { while(cin>>n) { memset(sum,0,sizeof(sum)); memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { cin>>a[i]; sum[i]=sum[i-1]+a[i]; } for(int i=1;i<=n;i++) dp[i][i]=0; for(int r=2;r<=n;r++){ for(int i=1;i<=n-r+1;i++){ int j=i+r-1; dp[i][j]=INF; for(int k=i;k<j;k++){ dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1]); } } } printf("%d\n",dp[1][n]); } return 0; }