Hdu 2513 Cake slicing【区间Dp】

Cake slicing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 252    Accepted Submission(s): 126

Problem Description

A rectangular cake with a grid of m*n unit squares on its top needs to be sliced into pieces. Several cherries are scattered on the top of the cake with at most one cherry on a unit square. The slicing should follow the rules below:
1.  each piece is rectangular or square;
2.  each cutting edge is straight and along a grid line;
3.  each piece has only one cherry on it;
4.  each cut must split the cake you currently cut two separate parts

For example, assume that the cake has a grid of 3*4 unit squares on its top, and there are three cherries on the top, as shown in the figure below.

One allowable slicing is as follows.

For this way of slicing , the total length of the cutting edges is 2+4=6.
Another way of slicing is 

In this case, the total length of the cutting edges is 3+2=5.

Give the shape of the cake and the scatter of the cherries , you are supposed to find
out the least total length of the cutting edges.

 

Input

The input file contains multiple test cases. For each test case:
The first line contains three integers , n, m and k (1≤n, m≤20), where n*m is the size of the unit square with a cherry on it . The two integers show respectively the row number and the column number of the unit square in the grid .
All integers in each line should be separated by blanks.

 

Output

Output an integer indicating the least total length of the cutting edges.

 

Sample Input

3 4 31 22 33 2

 

Sample Output

Case 1: 5

题目大意：

现在给你一个N*M的大蛋糕，上边一共有k个特殊物品，现在我们可以对一块蛋糕沿着格子线进行切割，现在希望将这个大蛋糕分成K块蛋糕，使得每一块蛋糕上边都有一个特殊物品，我们希望切割的长度最小，问这个最小长度。

思路：

求最小长度，又观察到数据范围，我们可以考虑暴力Dp去求解。

设定Dp【i】【j】【k】【l】表示现在蛋糕的剩余区间（矩形）为以（i，j）为左上角，（k，l）为右下角的情况的最小切割长度。

那么其状态转移方程也不难推出，分两种切割方法去暴力寻找切割点即可：

那么对于实现细节，我们可以直接记忆化搜索。

Ac代码：

#include<stdio.h>#include<iostream>#include<string.h>using namespace std;int n,m,tot;int a[25][25];int dp[25][25][25][25];int Dp(int i,int j,int k,int l){    if(dp[i][j][k][l]==-1)    {        int cnt=0;        for(int x=i;x<=k;x++)        {            for(int y=j;y<=l;y++)            {                if(a[x][y]==1)cnt++;            }        }        if(cnt<=1)        {            dp[i][j][k][l]=0;            return dp[i][j][k][l];        }        int minn=0x3f3f3f3f;        for(int tmp=i;tmp<k;tmp++)        {            minn=min(minn,Dp(i,j,tmp,l)+Dp(tmp+1,j,k,l)+(l-j+1));        }        for(int tmp=j;tmp<l;tmp++)        {            minn=min(minn,Dp(i,j,k,tmp)+Dp(i,tmp+1,k,l)+(k-i+1));        }        dp[i][j][k][l]=minn;        return dp[i][j][k][l];    }    else return dp[i][j][k][l];}int main(){    int kase=0;    while(~scanf("%d%d%d",&n,&m,&tot))    {        memset(dp,-1,sizeof(dp));        memset(a,0,sizeof(a));        for(int i=0;i<tot;i++)        {            int x,y;            scanf("%d%d",&x,&y);            a[x][y]=1;        }        Dp(1,1,n,m);        printf("Case %d: ",++kase);        printf("%d\n",dp[1][1][n][m]);    }}