Find a path

Find a path

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Description

Frog fell into a maze. This maze is a rectangle containing N rows and M 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−1, and Aavg is the average value of all Ai. The beauty of the path is (N+M–1) multiplies the variance of the values:(N+M−1)∑N+M−1i=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 T indicating the number of test cases (T≤50).

Each test case starts with a line containing two integers N and M (1≤N,M≤30). Each of the next N lines contains M 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”. X is the test case number starting from 1. Y is the minimum beauty value.

Sample Input

1

2 2

1 2

3 4

Sample Output

Case #1: 14

分析：

题意是：有一个N*M的网格，每个格子都有值，问从(1,1)到(N,M)(N+M−1)∑N+M−1i=1(Ai−Aavg)2 的最小值。



设：dp[i][j][k]表示到（i,j）且和为k的∑ni=1(Ai)2的最小值，所以递推公式为：

dp[i][j][k]=min(dp[i-1][j][k-m[i][j]]+m[i][j]*m[i][j],dp[i][j-1][k-m[i][j]]+m[i][j]*m[i][j],dp[i][j][k]);

最后d
[M][i]为到(N,M)的所有的可能的情况，在里面找最小的(N+M-1)*dp
[M][i]-i*i即可。

（感觉是动态规划+背包的思想）

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include<limits.h>
using namespace std;

const int maxn = 40;
long long  dp[maxn][maxn][1900];
int m[maxn][maxn];

long long min(long long a,long long b)
{
return a<b?a:b;
}

int main()
{
int T,N,M;
int cas=0;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&N,&M);
for(int i=1;i<=N;i++)
{
for(int j=1;j<=M;j++)
{
scanf("%d",&m[i][j]);
}
}
for(int i=0;i<=N;i++)
{
for(int j=0;j<=M;j++)
{
for(int k=0;k<=1800;k++)
{
dp[i][j][k]=LLONG_MAX;
}
}
}
dp[1][1][m[1][1]]=m[1][1]*m[1][1];
for(int i=1;i<=N;i++)
{
for(int j=1;j<=M;j++)
{
for(int k=m[i][j];k<=1800;k++)
{
if(dp[i-1][j][k-m[i][j]]<LLONG_MAX)
dp[i][j][k]=min(dp[i-1][j][k-m[i][j]]+m[i][j]*m[i][j],dp[i][j][k]);
if(dp[i][j-1][k-m[i][j]]<LLONG_MAX)
dp[i][j][k]=min(dp[i][j-1][k-m[i][j]]+m[i][j]*m[i][j],dp[i][j][k]);

}
}
}
long long ans = LLONG_MAX;
for(int i=0;i<=1800;i++)
{
if(dp
[M][i]!=LLONG_MAX)
ans = min((N+M-1)*dp
[M][i]-i*i,ans);
}
printf("Case #%d: %I64d\n",++cas,ans);
}
return 0;
}