hdu 5492 Find a path（公式推导，DP）

Find a path

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1409    Accepted Submission(s): 607

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

12 21 23 4

 

Sample Output

Case #1: 14

题意：求从（1,1）到（n,m）的一条路径，使得(N+M−1)∑N+M−1i=1(Ai−Aavg)2最小

思路：

把公式一步步化简后可以得到结果等于(n+m-1)*(A1²+A2²+...+A（n+m-1）²)-（A1+A2+...+An+m-1)² ）

由于数据量很小，只有30*30，并且有一条路径最多权值为60*30  所以我们可以用DP去做

dp[i][j][k]表示在(i,j)这个点的时候路径之和为k的时候A1²+A2²+...+Ak²的最小值

最后去枚举k，求出最小的即可

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>using namespace std;#define N 31#define INF 99999999int ma[N][N];int dp[N][N][N*60];int main(){    int T,n,m,tot=1;    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",&ma[i][j]);        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)               for(int k=0;k<=(n+m-1)*30;k++)                      dp[i][j][k]=INF;        dp[1][1][ma[1][1]]=ma[1][1]*ma[1][1];        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)                for(int k=0;k<=(n+m-1)*30;k++)                {                    if(i>1&&k>=ma[i][j]&&dp[i-1][j][k-ma[i][j]]!=INF) dp[i][j][k]=min(dp[i][j][k],dp[i-1][j][k-ma[i][j]]+ma[i][j]*ma[i][j]);                    if(j>1&&k>=ma[i][j]&&dp[i][j-1][k-ma[i][j]]!=INF) dp[i][j][k]=min(dp[i][j][k],dp[i][j-1][k-ma[i][j]]+ma[i][j]*ma[i][j]);                }        int ans=INF;        for(int i=1;i<=(n+m-1)*30;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",tot++,ans);    }    return 0;}