那些年没有AC的水题．．．

HDU 5492(ｄｐ

题目：一个数字矩阵，要求找到一条路径使得经过的数字的方差最小．

思路：稍微把式子变换一下，状态是走到当前格子和为ｋ时最小平方和．

/*
* @author:  Cwind
*/
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <functional>
#include <set>
#include <cmath>
using namespace std;
#define IOS std::ios::sync_with_stdio (false);std::cin.tie(0)
#define pb push_back
#define PB pop_back
#define bk back()
#define fs first
#define se second
#define sq(x) (x)*(x)
#define eps (1e-6)
#define IINF (1<<29)
#define LINF (1ll<<59)
#define INF (1000000000)
#define FINF (1e3)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> P;

const int maxsum=1900;
const int maxn=40;
int T,n,m;
int grid[maxn][maxn];
int dp[maxn][maxn][maxsum];
int h[maxn][maxn];
int pos[maxn][maxn][maxsum];
int inf;
int cas=0;
int main(){
    freopen("/home/files/CppFiles/in","r",stdin);
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    memset(dp,0x3f,sizeof dp);
    inf=dp[0][0][0];
    cin>>T;
    while(T--){
        cin>>n>>m;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                scanf("%d",&grid[i][j]);
            }
        }
        dp[0][0][grid[0][0]]=sq(grid[0][0]);
        pos[0][0][0]=grid[0][0];
        h[0][0]=1;
        int ans=inf;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                for(int k=0;k<h[i][j];k++){
                    int v=pos[i][j][k];
                    if(dp[i+1][j][v+grid[i+1][j]]==inf){
                        pos[i+1][j][h[i+1][j]++]=v+grid[i+1][j];
                        dp[i+1][j][v+grid[i+1][j]]=dp[i][j][v]+sq(grid[i+1][j]);
                    }else{
                        dp[i+1][j][v+grid[i+1][j]]=
                            min(dp[i][j][v]+sq(grid[i+1][j]),
                                dp[i+1][j][v+grid[i+1][j]]);
                    }
                    if(dp[i][j+1][v+grid[i][j+1]]==inf){
                        pos[i][j+1][h[i][j+1]++]=v+grid[i][j+1];
                        dp[i][j+1][v+grid[i][j+1]]=dp[i][j][v]+sq(grid[i][j+1]);
                    }else{
                        dp[i][j+1][v+grid[i][j+1]]=
                            min(dp[i][j][v]+sq(grid[i][j+1]),
                                dp[i][j+1][v+grid[i][j+1]]);
                    }
                    if(i==n-1&&j==m-1){
                        ans=min(ans,dp[i][j][v]*(n+m-1)-v*v);
                    }
                    dp[i][j][v]=inf;
                }
                h[i][j]=0;
            }
        }
        for(int i=0;i<=max(n,m);i++){
            for(int j=0;j<h[i][m];j++){
                int v=pos[i][m][j];
                dp[i][m][v]=inf;
            }
            h[i][m]=0;
            for(int j=0;j<h[n][i];j++){
                int v=pos[n][i][j];
                dp[n][i][v]=inf;
            }
            h[n][i]=0;
        }
        printf("Case #%d: %d\n",++cas,ans);
    }
    return 0;
}