HDU5492 Find a path （动态规划）

Find a path

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1866    Accepted Submission(s): 809

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

 

Source

2015 ACM/ICPC Asia Regional Hefei Online 

 

Recommend

wange2014

 

题意：走矩阵，每次只能往下或者往右，从坐上到右下。求方差。

方差有个公式是，D(x)=(E(x))^2-E(x^2)

直接dp，加上一维限制其中一个变量即可。

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <string>using namespace std;const int INF=1e9+7;const int MAXN=35;const int N=2000;int dp[MAXN][MAXN][N];int a[MAXN][MAXN];int b[MAXN][MAXN];int main(){    int t;    scanf("%d",&t);    int cas=1;    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]);                b[i][j]=a[i][j]*a[i][j];                for(int k=0;k<N;k++)                    dp[i][j][k]=INF;            }        }        dp[1][1][a[1][1]]=b[1][1];        for(int i=1;i<=n;i++){            for(int j=1;j<=m;j++){                if(i==1&&j==1)                    continue;                for(int k=0;k<N;k++){                    if(i>1&&k>=a[i][j]&&dp[i-1][j][k-a[i][j]]!=INF){                        dp[i][j][k]=min(dp[i-1][j][k-a[i][j]]+b[i][j],dp[i][j][k]);                    }                    if(j>1&&k>=a[i][j]&&dp[i][j-1][k-a[i][j]]!=INF){                        dp[i][j][k]=min(dp[i][j-1][k-a[i][j]]+b[i][j],dp[i][j][k]);                    }                }            }        }        int res=INF;        for(int k=0;k<N;k++){            if(dp[n][m][k]!=INF){                res=min(res,(n+m-1)*dp[n][m][k]-k*k);            }        }        printf("Case #%d: %d\n",cas++,res);    }}