poj棋盘分割（记忆化）

http://poj.org/problem?id=1191

黑书上P116 想了挺久 没想出来 想推出一公式来着 退不出来。。

想偏了  正解：递归

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0xfffffff
#define LL long long
int dp[20][10][10][10][10];
int aa[10][10],n;
int divide(int k,int a,int b,int c,int d)
{
    int i,j,minz = INF,g;
    if(dp[k][a][b][c][d])
    return dp[k][a][b][c][d];
    if(k==n)
    {
        int s = 0;
        for(i = a; i <= c ; i++)
            for(j = b ; j <= d ; j++)
            s+=aa[i][j];
        dp[k][a][b][c][d] = s*s;
        return s*s;
    }
    for(i = a+1 ; i <= c ; i++)
    {
        int s = 0;
        for(g = i ; g <= c ; g++)
        for(j = b ; j <= d ; j++)
        s+=aa[g][j];
        minz = min(minz,s*s+divide(k+1,a,b,i-1,d));
        s = 0;
        for(g = a ; g <= i-1 ; g++)
        for(j = b ; j <= d ; j++)
        s+=aa[g][j];
        minz = min(minz,s*s+divide(k+1,i,b,c,d));
    }
    for(i = b+1 ; i <= d ; i++)
    {
        int s = 0;
        for(g = a ; g <= c ; g++)
        for(j = i ; j <= d ; j++)
        s+=aa[g][j];
        minz = min(minz,s*s+divide(k+1,a,b,c,i-1));
        s = 0;
        for(g = a ; g <= c ; g++)
        for(j = b ; j <= i-1 ; j++)
        s+=aa[g][j];
        minz = min(minz,s*s+divide(k+1,a,i,c,d));
    }
    dp[k][a][b][c][d] = minz;
    return minz;
}
int main()
{
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        double s=0;
        memset(dp,0,sizeof(dp));
        int m = 8;
        n--;
        for(i = 1; i <= m ;i++)
            for(j = 1; j <= m ; j++)
            {
                scanf("%d",&aa[i][j]);
                s+=double(aa[i][j]);
            }
        s = s/double(n+1);
        int ans = divide(0,1,1,m,m);
        double an = 1.0/double(n+1)*double(ans)-s*s;
        printf("%.3lf
",sqrt(an));
    }
    return 0;
}

下面那个不是 贴错了

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#define N 5010
#define M 3010
#define INF 0xfffffff
using namespace std;
int dp[M][N],h[N];
int main()
{
    int i,j,n,m,g;
    scanf("%d%d",&m,&n);
    for(i = 1; i <= n ;i++)
    scanf("%d",&h[i]);
    for(i = 1; i <= m ;i++)
        for(j = 1; j <= n ;j++)
        dp[i][j] = INF;
    int o = INF;
    for(i = 2 ; i < n ;i++)
    {
        dp[1][i] = min(o,(h[i]-h[i-1])*(h[i]-h[i-1]));
        o = min(dp[1][i],o);
    }
    for(i = 2 ; i <= m ; i++)
    {
        int o = INF;
        for(j = 2*i ; j < n-(m-i)*3 ;j++)
        {
            if(j-2<(n-(m-i+1)*3))
            dp[i][j] = min(o,dp[i-1][j-2]+(h[j]-h[j-1])*(h[j]-h[j-1]));
            else
            dp[i][j] = min(o,dp[i-1][j-3]+(h[j]-h[j-1])*(h[j]-h[j-1]));
            o = min(o,dp[i][j]);
        }
    }
    printf("%d
",dp[m][n-1]);
    return 0;
}