codeChef----Maximum Score

You are given N integer sequences A₁, A₂, ..., A_N. Each of these sequences contains Nelements. You should pick N elements, one from each sequence; let's denote the element picked from sequence A_i by E_i. For each i (2 ≤ i ≤ N), E_i should be strictly greater than E_i-1.

Compute the maximum possible value of E₁ + E₂ + ... + E_N. If it's impossible to pick the elements E₁, E₂, ..., E_N, print -1 instead.

Input

• The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
• The first line of each test case contains a single integer N.
• N lines follow. For each valid i, the i-th of these lines contains N space-separated integers A_i1, A_i2, ..., A_iN denoting the elements of the sequence A_i.

Output

For each test case, print a single line containing one integer — the maximum sum of picked elements.

Constraints

• 1 ≤ T ≤ 10
• 1 ≤ N ≤ 700
• 1 ≤ sum of N in all test-cases ≤ 3700
• 1 ≤ A_ij ≤ 10⁹ for each valid i, j

Subtasks

Subtask #1 (18 points): 1 ≤ A_ij ≤ N for each valid i, j

Subtask #2 (82 points): original constraints

Example

Input:

1
3
1 2 3
4 5 6
7 8 9

Output:

18

Explanation

Example case 1: To maximise the score, pick 3 from the first row, 6 from the second row and 9 from the third row. The resulting sum is E₁+E₂+E₃ = 3+6+9 = 18.

Author:4★hruday968

Date Added:28-12-2017

Time Limit:1 secs

Source Limit:50000 Bytes

Languages:ADA, ASM, BASH, BF, C, CAML, CLOJ, CLPS, CPP 4.3.2, CPP 6.3, CPP14, CS2, D, ERL, FORT, FS, GO, HASK, ICK, ICON, JAVA, JS, kotlin, LISP clisp, LISP sbcl, LUA, NEM, NICE, NODEJS, PAS fpc, PAS gpc, PERL, PERL6, PHP, PIKE, PRLG, PYPY, PYTH, PYTH 3.5, RUBY, rust, SCALA, SCM chicken, SCM guile, SCM qobi, ST, swift, TCL, TEXT, WSPC

典型的动态规划问题，直接DP就能实现。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
typedef long long ll;
using namespace std;
ll n;
ll a[710][710];
ll dp[710][710];//使用long long 形式，若用int，则会炸点数据 
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld",&n);
		for(ll i=0;i<n;i++)
		{
			for(ll j=0;j<n;j++) scanf("%lld",&a[i][j]);
		}
		for(ll i=0;i<n;i++) sort(a[i],a[i]+n);
		memset(dp,0,sizeof(dp));
		for(ll i=0;i<n;i++) dp[0][i]=a[0][i];
		for(ll i=1;i<n;i++)
		{
			for(ll j=0;j<n;j++)
			{
				ll t=a[i][j];
				for(ll k=0;k<n;k++)
				{
					if(t>a[i-1][k] && dp[i-1][k]>0) dp[i][j]=max(dp[i][j],dp[i-1][k]+t);
				}
			}
		}
		ll maxx=0;
		for(ll i=0;i<n;i++) maxx=max(maxx,dp[n-1][i]);
		if(maxx==0) printf("-1\n");
		else printf("%lld\n",maxx);//该方法因为n最大为700，测试用例子最大时间为980ms，在1000ms限制之内。
	}
	return 0;
}