C - Decreasing Heights

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这是一个关于解决三维沙盒游戏中从起点到终点路径优化的问题。给定一个n×m的世界地图矩阵，每格的高度为ai,j，并且只能向下或向右移动，且移动时高度必须递增。允许在开始前对地图进行操作，降低任意格子的高度。任务是找到最小的操作次数，使得从(1,1)到(n,m)存在一条合法路径。输入包含多个测试用例，每个用例包含地图尺寸和高度数据，输出每个用例所需的最小操作数。给出的示例展示了如何计算并输出答案。

链接：https://vjudge.net/contest/399019#problem/C

You are playing one famous sandbox game with the three-dimensional world. The map of the world can be represented as a matrix of size n×mn×m, where the height of the cell (i,j)(i,j) is ai,jai,j.

You are in the cell (1,1)(1,1) right now and want to get in the cell (n,m)(n,m). You can move only down (from the cell (i,j)(i,j) to the cell (i+1,j)(i+1,j)) or right (from the cell (i,j)(i,j) to the cell (i,j+1)(i,j+1)). There is an additional restriction: if the height of the current cell is xx then you can move only to the cell with height x+1x+1.

Before the first move you can perform several operations. During one operation, you can decrease the height of any cell by one. I.e. you choose some cell (i,j)(i,j) and assign (set) ai,j:=ai,j−1ai,j:=ai,j−1. Note that you can make heights less than or equal to zero. Also note that you can decrease the height of the cell (1,1)(1,1).

Your task is to find the minimum number of operations you have to perform to obtain at least one suitable path from the cell (1,1)(1,1) to the cell (n,m)(n,m). It is guaranteed that the answer exists.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases. Then tt test cases follow.

The first line of the test case contains two integers nn and mm (1≤n,m≤1001≤n,m≤100) — the number of rows and the number of columns in the map of the world. The next nn lines contain mm integers each, where the jj-th integer in the ii-th line is ai,jai,j (1≤ai,j≤10151≤ai,j≤1015) — the height of the cell (i,j)(i,j).

It is guaranteed that the sum of nn (as well as the sum of mm) over all test cases does not exceed 100100 (∑n≤100;∑m≤100∑n≤100;∑m≤100).

Output

For each test case, print the answer — the minimum number of operations you have to perform to obtain at least one suitable path from the cell (1,1)(1,1) to the cell (n,m)(n,m). It is guaranteed that the answer exists.

Example

Input

5
3 4
1 2 3 4
5 6 7 8
9 10 11 12
5 5
2 5 4 8 3
9 10 11 5 1
12 8 4 2 5
2 2 5 4 1
6 8 2 4 2
2 2
100 10
10 1
1 2
123456789876543 987654321234567
1 1
42

Output

9
49
111
864197531358023
0

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll maxn=1e3+7;
const ll mod=0x3f3f3f3f3f3f3f3f;
ll t,n,m,k,s,ans,sum;
char x[maxn];
ll dp[maxn][maxn];
ll a[maxn][maxn];
ll b[maxn][maxn];
ll f(int p,int q)
{
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            b[i][j]=a[i][j]-a[p][q];
            dp[i][j]=mod;
        }
    }
    if(b[n][m]<0||b[1][1]<0)
        return mod;
    dp[1][1]=b[1][1];
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(b[i][j]<0)
                continue;
			if(i>1&&b[i-1][j]>=0)
			dp[i][j]=min(dp[i][j],dp[i-1][j]+b[i][j]);
			if(j>1&&b[i][j-1]>=0)
			dp[i][j]=min(dp[i][j],dp[i][j-1]+b[i][j]);
        }
    }
    //cout<<dp[n][n]<<endl;
    return dp[n][m];
}

int main()
{
    cin>>t;
    while(t--)
    {
        scanf("%lld %lld",&n,&m);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                scanf("%lld",&a[i][j]);
                a[i][j]-=i+j-2;
            }
        }
        ans=mod;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                s=f(i,j);
                ans=min(ans,s);
            }
        }
        cout<<ans<<'\n';
    }
    return 0;
}