DP HDOJ 5492 Find a path

题目传送门

题意：从(1, 1)走到(n, m)，每次往右或往下走，问(N+M−1)∑(Ai−Aavg)2 的最小值

分析：展开式子得到(N+M−1)∑(Ai2) - (∑(Ai))2的最小值。用普通的搜索要不超时要不爆内存，用dp。注意到和的值很小，最多59*30，所以dp[i][j][k]表示当走到(i, j)点时和为k的最小的平方和，两个方向转移。

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/28 星期一 08:16:33
* File Name     :I.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 33;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
int dp[N][N][N*2*N];
int a[N][N];

int main(void)    {
    int T, cas = 0;  scanf ("%d", &T);
    while (T--) {
        int n, m;   scanf ("%d%d", &n, &m);
        for (int i=1; i<=n; ++i)    {
            for (int j=1; j<=m; ++j)    {
                scanf ("%d", &a[i][j]);
            }
        }
        int S = 59 * 30;
        memset (dp, INF, sizeof (dp));
        dp[1][1][a[1][1]] = a[1][1] * a[1][1];
        for (int i=1; i<=n; ++i)    {
            for (int j=1; j<=m; ++j)    {
                for (int k=0; k<=S; ++k)    {
                    int &u = dp[i][j][k];
                    if (u == INF)   continue;
                    if (i + 1 <= n)  {
                        int &v = dp[i+1][j][k+a[i+1][j]];
                        v = min (v, u + a[i+1][j] * a[i+1][j]);
                    }
                    if (j + 1 <= m)  {
                        int &v = dp[i][j+1][k+a[i][j+1]];
                        v = min (v, u + a[i][j+1] * a[i][j+1]);
                    }
                }
            }
        }
        int ans = INF;
        for (int i=0; i<=S; ++i)    {
            if (dp[n][m][i] == INF) continue;
            ans = min (ans, (n + m - 1) * dp[n][m][i] - i * i);
        }
        printf ("Case #%d: %d\n", ++cas, ans);
    }

    return 0;
}