hdoj5492Find a path【dp】

Find a path

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1173    Accepted Submission(s): 511


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

1
2 2
1 2
3 4

 

Sample Output

Case #1: 14

 

Source

2015 ACM/ICPC Asia Regional Hefei Online

 

题意：给一个矩形迷宫每个格子上有一个值Ai求从左上角走到右下角使得(N+M−1)
∑N+M−1i=1(Ai−Aavg)2值最小观察此式和方差的计算公式类似因此可利用
方差的计算公式简化此式为(N+M−1)∑A2i−(∑Ai)2
参考博客：http://blog.csdn.net/u012860063/article/details/48790895

/*
dp[i][j][k]在第i行第j个格子处所经过的路径和为k时格子的平方和的最小值
*/
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;
int map[35][35];
int dp[35][35][2010];
int main()
{
int t,n,m,i,j,test=1,k;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(i=1;i<=n;++i){
for(j=1;j<=m;++j){
scanf("%d",&map[i][j]);
}
}
memset(dp,0x3f,sizeof(dp));
dp[1][0][0]=dp[0][1][0]=0;
for(i=1;i<=n;++i){
for(j=1;j<=m;++j){
for(k=0;k<=1800;++k){
dp[i][j][k+map[i][j]]=min(dp[i][j][k+map[i][j]],min(dp[i-1][j][k],dp[i][j-1][k])+map[i][j]*map[i][j]);
}
}
}
int ans=inf;
for(k=0;k<=1800;++k){
if(dp
[m][k]!=inf)
ans=min(ans,dp
[m][k]*(n+m-1)-k*k);
}
printf("Case #%d: %d\n",test++,ans);
}
return 0;
}

i=1∑N+M−1i=1)