poj-1191- 棋盘分割dp

做法：

设立数组f[x1][y1][x2][y2][k]表示在矩形x1y1,x2y2中还需切割k-1次；

逐步递推即可。

 

#include<stdio.h>

#include<math.h>

#include<iostream>

#include<string.h>

#include<algorithm>

#include<queue>

#include<stack>

#include<map>

#include<string>

#include<stdlib.h>

#define INF_MAX 0x7fffffff

#define INF 999999

#define max3(a,b,c) (max(a,b)>c?max(a,b):c)

#define min3(a,b,c) (min(a,b)<c?min(a,b):c)

#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

struct node

{

    int u;

    int v;

    int w;

    bool friend operator < (node a, node b){

        return a.w < b.w;

    }

}edge[1001];

int gcd(int n,int m){if(n<m) swap(n,m);return n%m==0?m:gcd(m,n%m);}

int lcm(int n,int m){if(n<m) swap(n,m);return n/gcd(n,m)*m;}

int f[10][10][10][10][21];

int p[10][10][10][10];

int a[10][10];

int init(int x1,int y1,int x2,int y2)

{

    if(p[x1][y1][x2][y2]>=0)return p[x1][y1][x2][y2];

    int sum;

    sum=0;

    int i,j;

    for(i=x1;i<=x2;i++)

    {

        for(j=y1;j<=y2;j++)

        {

            sum+=a[i][j];

        }

    }

    p[x1][y1][x2][y2]=sum*sum;

   // printf("-----%d %d %d %d %d\n",x1,y1,x2,y2,sum*sum);

    return sum*sum;

}

int dos(int x1,int y1,int x2,int y2,int k)

{



    int i;

    if(k==1)return init(x1,y1,x2,y2);

    if(f[x1][y1][x2][y2][k]>=0)return f[x1][y1][x2][y2][k];

    int mins=99999999;

    for(i=x1;i<x2;i++)

    {

        mins=min3(mins,(dos(i+1,y1,x2,y2,k-1)+init(x1,y1,i,y2)),(init(i+1,y1,x2,y2)+dos(x1,y1,i,y2,k-1)));

    }

    for(i=y1;i<y2;i++)

    {

        mins=min3(mins,(dos(x1,i+1,x2,y2,k-1)+init(x1,y1,x2,i)),(init(x1,i+1,x2,y2)+dos(x1,y1,x2,i,k-1)));

    }

    f[x1][y1][x2][y2][k]=mins;

    //printf("%d %d %d %d %d %d\n",x1,y1,x2,y2,k,mins);

    return mins;

}

int main()

{

    int n,i,j,ans;

    scanf("%d",&n);

    double x;

    x=0;

    mem(f,-1);

    mem(p,-1);

    for(i=1;i<=8;i++)

    {

        for(j=1;j<=8;j++)

            scanf("%d",&a[i][j]);

    }

    x=1.0*x/n;

    x=x*x;

    ans=dos(1,1,8,8,n);

    x=(sqrt(1.0*init(1,1,8,8))/n);

    printf("%.3f\n",sqrt(1.0*ans/n-x*x));

    return 0;

}