动态规划——Cake slicing

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

题意：给你一个n×m的网格，然后有k个樱桃的坐标，让你去切成几个矩形方块，要求每个方块上面都有个樱桃，求最短的切割距离。

思路：

记忆化搜索，用dp[sx][sy][ex][ey]表示左上角为（sx，sy）和右下角为（ex， ey）的这块矩形上所需的最短距离。

若这块矩形上无点就表示无穷大，有一个点就表示为0，多余一个点就继续分割

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;const int maxn=30;int n, m , k;int dp[maxn][maxn][maxn][maxn];int pl[maxn][maxn];int check(int sx, int sy, int ex, int ey)    //判断矩形上有无点{    int num=0;    for(int i=sx+1; i<=ex; i++)        for(int j=sy+1; j<=ey; j++)            if(pl[i][j])            {                num++;                if(num==2)                    return num;            }    return num;}int dfs(int sx, int sy, int ex, int ey){    if(dp[sx][sy][ex][ey]!=-1)        return dp[sx][sy][ex][ey];    if(check(sx, sy, ex,ey)==0)        return dp[sx][sy][ex][ey]=INF;    if(check(sx, sy, ex, ey)==1)        return dp[sx][sy][ex][ey]=0;    int ans=INF;    for(int i=sy+1; i<ey; i++)    //竖着割        ans=min(ans, dfs(sx, i, ex, ey)+dfs(sx, sy, ex, i)+ex-sx);    for(int i=sx+1; i<ex; i++)    //横着割        ans=min(ans, dfs(i, sy, ex, ey)+dfs(sx, sy, i, ey)+ey-sy);    return dp[sx][sy][ex][ey]=ans;}int main(){    int cas=1;    while(~scanf("%d%d%d", &m, &n, &k))    {        int x, y;        memset(pl, 0, sizeof(pl));        memset(dp, -1, sizeof(dp));        for(int i=0; i<k; i++)    //将樱桃标记        {            scanf("%d%d", &x, &y);            pl[x][y]=1;        }        printf("Case %d: %d\n",cas++,dfs(0,0,m,n));    }    return 0;}