HDU 5492 Find a path (dp)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5492

每个数字只能走下或者走右，问你方差最小多少？

方差 = (n + m - 1) * sum(a[i]^2) + sum(a[i])

我们可以定义dp[i][j][k]为(i, j)点和为k的最小平方和，k最大也就(n-m+1)*30大小

那么可以dp[i][j][k - a[i][j]] = min(dp[i][j - 1][k], dp[i - 1][j][k]) + a[i][j]*a[i][j];

保证转移最优的话，那么需要初始化dp[i][j][k] = inf;

//#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 1e5 + 5;
int dp[35][35][1805], a[35][35], inf = 1e6;

int main()
{
    int t, n, m;
    scanf("%d", &t);
    for(int ca = 1; ca <= t; ++ca) {
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= m; ++j) {
                scanf("%d", &a[i][j]);
                for(int k = 0; k <= 1800; ++k) {
                    dp[i][j][k] = inf;
                }
            }
        }
        int sum = 0;
        for(int i = 1; i <= n; ++i) {
            sum += a[i][1];
            dp[i][1][sum] = dp[i - 1][1][sum - a[i][1]] + a[i][1]*a[i][1];
        }
        sum = 0;
        for(int i = 1; i <= m; ++i) {
            sum += a[1][i];
            dp[1][i][sum] = dp[1][i - 1][sum - a[1][i]] + a[1][i]*a[1][i];
        }
        for(int i = 2; i <= n; ++i) {
            for(int j = 2; j <= m; ++j) {
                for(int k = a[i][j]; k <= 1800; ++k) {
                    dp[i][j][k] = min(dp[i - 1][j][k - a[i][j]], dp[i][j - 1][k - a[i][j]]) + a[i][j]*a[i][j];
                }
            }
        }
        int ans = inf;
        for(int k = 0; k <= 1800; ++k) {
            if(dp[n][m][k] < inf) {
                ans = min(ans, dp[n][m][k] * (n + m - 1) - k*k);
            }
        }
        printf("Case #%d: %d\n", ca, ans);
    }
    return 0;
}