HDU 5492 Find a path

Find a path

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 930    Accepted Submission(s): 410

Problem Description

Frog fell into a maze. This maze is a rectangle containing  rows and  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 , and  is the average value of all . The beauty of the path is  multiplies the variance of the values:
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  indicating the number of test cases ().
Each test case starts with a line containing two integers  and  (). Each of the next  lines contains  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”.  is the test case number starting from 1.  is the minimum beauty value.

 

Sample Input

1 2 2 1 2 3 4

 

Sample Output

Case #1: 14

 

Source

2015 ACM/ICPC Asia Regional Hefei Online

 

这个是合肥网络赛的一道题目，知道是dp但是当时没做出来

题目大体意思是有m*n的棋盘 从左上角走到右下角 求路径上经过的数字的方差最小

当时想的是 f[i][j][k]表示走到i,j 和为k的时候的最小值

其实就差一步就过了

当时想的是 要统计出来 经过i,j走到m,n可行的取值k

因为担心有一种情况 从1,1 经过i,j 到达 m,n 路径上的数字的和取不到k 但是用k当作和求方差导致错误

其实这种情况是不存在的

如果把平均值ave看作变量 x1,x2,...,xn看作常数

写出来方差的公式

整理一下发现是关于ave的二次方程

当ave=(x1+x2+...+xn)/n时取最小值

所以不会出现 用不是路径和的k得到更小的值

所以枚举k 直接做就行

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
int f[31][31];
int s[31][31];
int m,n,x,y;
int t;
int ans;
int main()
{
    scanf("%d",&t);
    for(int it=1;it<=t;it++)
    {
        scanf("%d%d",&m,&n);
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&s[i][j]);
        int w=(m+n-1);
        ans=0x3f3f3f3f;
        for(int lim=0;lim<=w*30;lim++)
        {
            memset(f,0x3f,sizeof(f));
            for(int i=1;i<=m;i++)
                for(int j=1;j<=n;j++)
                {
                    if(i==1 &&j==1)
                        f[i][j]=(w*s[i][j]-lim)*(w*s[i][j]-lim);
                    if(i>1)
                        f[i][j]=min(f[i][j],f[i-1][j]+(w*s[i][j]-lim)*(w*s[i][j]-lim));
                    if(j>1)
                        f[i][j]=min(f[i][j],f[i][j-1]+(w*s[i][j]-lim)*(w*s[i][j]-lim));
                }
            ans=min(ans,f[m][n]);
        }
        printf("Case #%d: %d
",it,ans/w);
    }
    return 0;
}

另外一种方法是看了别人的题解明白的

http://blog.csdn.net/queuelovestack/article/details/48768275

我摘抄过来

/************************************************************************************

首先，因为公式涉及权值的平均数，不妨将其化简一下

N+M-1的值是固定的，所以只跟和有关

又因为Ai的值是不超过30的，而N+M-1个Ai相加最多不超过1800，故我们可以用个三维dp来解此题,dp[i][j][k]表示从点(1,1)到点(i,j)的值为k时，的值是dp[i][j][k]，那么我们最后只需求出min((M+N-1)dp[i][j][k]-k*k)即可

代码如下

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int f[31][31][61*30+1];
int s[31][31];
int t,m,n;

int main()
{
    scanf("%d",&t);
    for(int it=1;it<=t;it++)
    {
        scanf("%d%d",&m,&n);
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&s[i][j]);
                for(int k=0;k<=(m+n-1)*30;k++)
                    f[i][j][k]=1e8;
            }
        f[1][1][s[1][1]]=s[1][1]*s[1][1];
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
                for(int k=0;k<=(m+n-1)*30;k++)
                {
                    if(i==1 && j==1 &&k==s[i][j])
                        f[i][j][k]=s[i][j]*s[i][j];
                    if(k>=s[i][j])
                    {
                        if(i>1)
                            f[i][j][k]=min(f[i][j][k],f[i-1][j][k-s[i][j]]+s[i][j]*s[i][j]);
                        if(j>1)
                            f[i][j][k]=min(f[i][j][k],f[i][j-1][k-s[i][j]]+s[i][j]*s[i][j]);
                    }
                }
        int ans=1e9;
        int temp=(m+n-1);
        for(int k=0;k<=temp*30;k++)
            if(f[m][n][k]!=1e8)
            ans=min(ans,temp*f[m][n][k]-k*k);
        printf("Case #%d: %d
",it,ans);
    }
    return 0;
}