题干：
Frog fell into a maze. This maze is a rectangle containing NN rows and MM columns. Each grid in this maze contains a number, which is called the magic value. Frog now stays at grid (1, 1), and he wants to go to grid (N, M). For each step, he can go to either the grid right to his current location or the grid below his location. Formally, he can move from grid (x, y) to (x + 1, y) or (x, y +1), if the grid he wants to go exists.
Frog is a perfectionist, so he'd like to find the most beautiful path. He defines the beauty of a path in the following way. Let’s denote the magic values along a path from (1, 1) to (n, m) as A1,A2,…AN+M−1A1,A2,…AN+M−1, and AavgAavg is the average value of all AiAi. The beauty of the path is (N+M–1)(N+M–1) multiplies the variance of the values:(N+M−1)∑N+M−1i=1(Ai−Aavg)2(N+M−1)∑i=1N+M−1(Ai−Aavg)2
In Frog's opinion, the smaller, the better. A path with smaller beauty value is more beautiful. He asks you to help him find the most beautiful path.
Input
The first line of input contains a number TT indicating the number of test cases (T≤50T≤50).
Each test case starts with a line containing two integers NN and MM (1≤N,M≤301≤N,M≤30). Each of the next NN lines contains MM non-negative integers, indicating the magic values. The magic values are no greater than 30.
Output
For each test case, output a single line consisting of “Case #X: Y”. XX is the test case number starting from 1. YY is the minimum beauty value.
Sample Input
1
2 2
1 2
3 4
Sample Output
Case #1: 14
题目大意：
在一个N*M的数字方格中，寻找一条从左上角(1,1)到右下角(n,m)的路径，每次只能往右走或往下走。
使得路径上的数字方差（或者说
这个式子）最小。（n,m,a[i][j]均<=30）
解题报告：
如果说是方差的话更加不好思考，还是直接看上面这个式子，不难化简出：原式
。
然后本来想直接贪心这个式子最小的情况下转移，但是无法证明正确性。
其实我们不妨多一维状态，记录dp[i][j][k]代表从(1,1)走到(i,j)且路径的和为k的最小平方和。因为要求式子的值最小，所以在原式路径和相同的情况下被减数越小越好，又因为(i,j)不论从哪里转移过来，此时的(n+m-1)肯定为(i+j-1)，是个定值，所以只需要维护平方和最小值即可。注意初始化状态的时候不是dp[0][0][0]了。。通过画二维矩阵就容易看出初始化的状态是哪些。
AC代码：
using namespace std;
typedef pair<int,int> PII;
const int MAX = 55 + 5;
ll a[MAX][MAX];
ll dp[55][55][2222];
int n,m;
int main()
{
int T,iCase=0;
cin>>T;
while(T--) {
scanf("%d%d",&n,&m);
int sum = 0;
for(int i = 1; i<=n; i++) {
for(int j = 1; j<=m; j++) {
scanf("%lld",&a[i][j]);sum += a[i][j];
}
}
for(int i = 0; i<=n; i++) {
for(int j = 0; j<=m; j++) for(int k = 0; k<=2000; k++) dp[i][j][k] = 1e12;
}
dp[0][1][0]=dp[1][0][0]=0;
for(int i = 1; i<=n; i++) {
for(int j = 1; j<=m; j++) {
for(int k = a[i][j]; k<=2000; k++) {
dp[i][j][k] = min(dp[i-1][j][k-a[i][j]]+a[i][j]*a[i][j],dp[i][j-1][k-a[i][j]]+a[i][j]*a[i][j]);
}
}
}
ll ans = 0x3f3f3f3f;
for(int i = 0; i<=2000; i++) {
ans = min(1LL*ans,(n+m-1)*dp[n][m][i]-i*i);
}
printf("Case #%d: %lld\n",++iCase,ans);
}
return 0 ;
}
/*
1
2 2
1 2
3 4
*/
