Problem:
Score : 1200 points

Problem Statement

We have a weighted directed graph with N vertices numbered 0 to N-1.

The graph initially has N-1 edges. The i-th edge (0 \leq i \leq N-2) is directed from Vertex i to Vertex i+1 and has a weight of 0.

Snuke will now add a new edge (i → j) for every pair i, j (0 \leq i,j \leq N-1,\ i \neq j).
The weight of the edge will be -1 if i < j, and 1 otherwise.

Ringo is a boy. A negative cycle (a cycle whose total weight is negative) in a graph makes him sad.
He will delete some of the edges added by Snuke so that the graph will no longer contain a negative cycle.
The cost of deleting the edge (i → j) is A_{i,j}. He cannot delete edges that have been present from the beginning.

Find the minimum total cost required to achieve Ringo's objective.

Constraints

3 \leq N \leq 500

1 \leq A_{i,j} \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_{0,1} A_{0,2} A_{0,3} \cdots A_{0,N-1}
A_{1,0} A_{1,2} A_{1,3} \cdots A_{1,N-1}
A_{2,0} A_{2,1} A_{2,3} \cdots A_{2,N-1}
\vdots
A_{N-1,0} A_{N-1,1} A_{N-1,2} \cdots A_{N-1,N-2}

Output

Print the minimum total cost required to achieve Ringo's objective.

Sample Input 1

3
2 1
1 4
3 3

Sample Output 1

2

If we delete the edge (0 → 1) added by Snuke, the graph will no longer contain a negative cycle.
The cost will be 2 in this case, which is the minimum possible.

Sample Input 2

4
1 1 1
1 1 1
1 1 1
1 1 1

Sample Output 2

2

If we delete the edges (1 → 2) and (3 → 0) added by Snuke, the graph will no longer contain a negative cycle.
The cost will be 1+1=2 in this case, which is the minimum possible.

Sample Input 3

10
190587 2038070 142162180 88207341 215145790 38 2 5 20
32047998 21426 4177178 52 734621629 2596 102224223 5 1864
41 481241221 1518272 51 772 146 8805349 3243297 449
918151 126080576 5186563 46354 6646 491776 5750138 2897 161
3656 7551068 2919714 43035419 495 3408 26 3317 2698
455357 3 12 1857 5459 7870 4123856 2402 258
3 25700 16191 102120 971821039 52375 40449 20548149 16186673
2 16 130300357 18 6574485 29175 179 1693 2681
99 833 131 2 414045824 57357 56 302669472 95
8408 7 1266941 60620177 129747 41382505 38966 187 5151064

Sample Output 3

2280211

Solution in C++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

const int maxn=505;
int n;
int w[maxn][maxn]={0},p[maxn][maxn]={0},tot[maxn][maxn]={0},dp[maxn][maxn]={0};
int ans=2147483647;

inline int getsum(int u,int v,int x,int y){
	return p[v][y]-(x==0?0:p[v][x-1])-(u==0?0:p[u-1][y])+(u==0 || v==0?0:p[u-1][x-1]);
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(i!=j)  scanf("%d",&w[i][j]);
			
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			p[i][j]=p[i][j-1]+p[i-1][j]-p[i-1][j-1]+w[i][j];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			tot[i][j]=tot[i][j-1]+tot[i-1][j]-tot[i-1][j-1]+(i<j?w[i][j]:0);
			
	for(int i=1;i<=n;i++){
		dp[i][1]=tot[i][i]-tot[i][0]-tot[0][i]+tot[0][0];
		for(int j=2;j<=i;j++){
			dp[i][j]=0x3f3f3f3f;
			for(int k=1;k<=j-1;k++)
				dp[i][j]=min(dp[i][j],dp[j-1][k]+(tot[i][i]-tot[i][j-1]-tot[j-1][i]+tot[j-1][j-1])+getsum(j,i,0,k-1));
		}
	}
	
	for(int i=1;i<=n;i++)  ans=min(ans,dp[n][i]);
	printf("%d",ans);
	
	return 0;
}