Hdu 5492 Find a path【Dp】

最新推荐文章于 2019-11-18 19:56:01 发布

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最新推荐文章于 2019-11-18 19:56:01 发布

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分类专栏： dp 思维稍带数学的思维题文章标签： Hdu 5492

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本文介绍了一种求解迷宫中从起点到终点最短路径的算法，通过定义路径的美观度为路径上各点数值的方差，并利用动态规划的思想进行求解。

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Find a path

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1759 Accepted Submission(s): 770

Problem 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

Sample Output

Case #1: 14

题目大意：

给你一个N*M的矩阵，我们从左上角走到右下角，走到的位子上的数字必须拿取，而且每一步只能向右走或者向下走。

让你求公式的最小值。

思路：

我们分解开公式，不难化简式子变成：（n+m-1）*ΣAi^2-(ΣAi)^2；

那么我们利用背包思想，我们设定ΣAi为背包容量，我们希望价值最小，所以我们希望ΣAi^2越小，所以我们设定Dp【i】【j】【k】表示到点（i，j）处，ΣAi的值为k能够获得的最小的ΣAi^2的值。

那么其状态转移方程不难写出：

然后最终暴力枚举所有情况维护最小值即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int a[45][45];
int dp[35][35][60*32*2];
int main()
{
    int t;
    int kase=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]);
                for(int k=0;k<=(n+m)*60;k++)
                {
                    dp[i][j][k]=0x3f3f3f3f;
                }
            }
        }
        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<=(n+m)*60;k++)
                {
                    if(i==1&&j==1)continue;
                    if(i-1>=1&&k-a[i][j]>=0&&dp[i-1][j][k-a[i][j]]!=0x3f3f3f3f)dp[i][j][k]=min(dp[i][j][k],dp[i-1][j][k-a[i][j]]+a[i][j]*a[i][j]);
                    if(j-1>=1&&k-a[i][j]>=0&&dp[i][j-1][k-a[i][j]]!=0x3f3f3f3f)dp[i][j][k]=min(dp[i][j][k],dp[i][j-1][k-a[i][j]]+a[i][j]*a[i][j]);
                }
            }
        }
        int output=0x3f3f3f3f;
        for(int k=0;k<=(n+m)*60;k++)
        {
            if(dp[n][m][k]!=0x3f3f3f3f)
            {
               output=min(output,(n+m-1)*(dp[n][m][k])-k*k);
            }
        }
        printf("Case #%d: %d\n",++kase,output);
    }
}

(N+M−1)∑N+M−1i=1(Ai−Aavg)2