题意：一个矩形蛋糕上有好多个樱桃，现在要做的就是切割最少的距离，切出矩形形状的小蛋糕，让每个蛋糕上都有一个樱桃，问最少切割距离是多少。
析：很容易知道是记忆化搜索，我们用dp[u][d][l][r]来表示，上界是u，下界是d，左边是l，右边是r，然后不断切割，不过要注意切的时候是按缝隙切，
缝隙多一条，那么我们可以补上一条，用0来补齐，然后就进行计算就好。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 100000000000000000;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 20 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
inline LL Max(LL a, LL b){ return a < b ? b : a; }
inline LL Min(LL a, LL b){ return a > b ? b : a; }
inline int Max(int a, int b){ return a < b ? b : a; }
inline int Min(int a, int b){ return a > b ? b : a; }
bool cake[maxn][maxn];
int dp[maxn][maxn][maxn][maxn]; int cal(int u, int d, int l, int r){
int cnt = 0;
for(int i = u+1; i <= d; ++i)
for(int j = l+1; j <= r; ++j){
if(cake[i][j]) ++cnt;
if(cnt >= 2) return 2;
}
return cnt;
} int DP(int u, int d, int l, int r){
int &ans = dp[u][d][l][r];
if(ans >= 0) return ans;
int num = cal(u, d, l, r);
if(1 == num) return ans = 0;
if(!num) return ans = INF;
ans = INF;
for(int i = u+1; i < d; ++i)
ans = Min(ans, DP(u, i, l, r) + DP(i, d, l, r) + r - l);
for(int i = l+1; i < r; ++i)
ans = Min(ans, DP(u, d, l, i) + DP(u, d, i, r) + d - u);
return ans;
} int main(){
int k, kase = 0;
while(scanf("%d %d %d", &n, &m, &k) == 3){
memset(cake, false, sizeof cake);
for(int i = 0; i < k; ++i){
int x, y;
scanf("%d %d", &x, &y);
cake[x][y] = true;
}
memset(dp, -1, sizeof dp);
printf("Case %d: %d\n", ++kase, DP(0, n, 0, m));
}
return 0;
}