HDU - 5492 Find a path（方差公式+dp）

Find a path

 

Frog fell into a maze. This maze is a rectangle containing (N+M−1)∑i=1N+M−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. 

InputThe first line of input contains a number Y is the minimum beauty value.Sample Input

1
2 2
1 2
3 4

Sample Output

Case #1: 14




将公式变形得：（n+m-1）*ΣAi^2-(ΣAi)^2
dp求出每种和的最小的平方和，最后找出满足公式的最小解。

#include<bits/stdc++.h>
#define MAX 31
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

int a[MAX][MAX];
int dp[MAX][MAX][1801];

int main()
{
    int tt=0,t,n,m,i,j,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",&a[i][j]);
            }
        }
        memset(dp,INF,sizeof(dp));
        dp[1][0][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++){
                    if(k+a[i][j]<=1800) dp[i][j][k+a[i][j]]=min(dp[i][j][k+a[i][j]],min(dp[i-1][j][k],dp[i][j-1][k])+a[i][j]*a[i][j]);
                }
            }
        }
        int ans=INF;
        for(i=0;i<=1800;i++){
            if(dp[n][m][i]!=INF){
                ans=min(ans,(n+m-1)*dp[n][m][i]-i*i);
            }
        }
        printf("Case #%d: %d\n",++tt,ans);
    }
    return 0;
}