题意：有一个n行m列(1<=n, m<=20)的网格蛋糕上有一些樱桃。每次可以用一刀沿着网格线把蛋糕切成两块，并且只能够直切不能拐弯。要求最后每一块蛋糕上恰好有一个樱桃，且切割线总长度最小。
分析：dp[up][down][left][right]表示上下左右界分别为up,down,left,right的蛋糕，为了使最后每一块蛋糕上恰好有一个樱桃，切割线的最小总长度。
#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 20 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int pic[MAXN][MAXN];
int dp[MAXN][MAXN][MAXN][MAXN];
int getNum(int up, int down, int left, int right){//求蛋糕上的樱桃数
int ans = 0;
for(int i = up; i <= down; ++i){
for(int j = left; j <= right; ++j){
if(pic[i][j]) ++ans;
}
}
return ans;
}
int dfs(int num, int up, int down, int left, int right){//num蛋糕上的樱桃数
if(dp[up][down][left][right] != -1) return dp[up][down][left][right];
if(num == 1) return dp[up][down][left][right] = 0;
int ans = INT_INF;
for(int i = up; i < down; ++i){//枚举分割线,i为第i行下面的分隔线，横切
int tmp = getNum(up, i, left, right);//切割完上半部分蛋糕上的樱桃数
if(tmp >= 1 && tmp < num){//被切割成的两块蛋糕上都有樱桃
int x = dfs(tmp, up, i, left, right);
int y = dfs(num - tmp, i + 1, down, left, right);
ans = Min(ans, x + y + right - left + 1);
}
}
for(int i = left; i < right; ++i){//竖切
int tmp = getNum(up, down, left, i);
if(tmp >= 1 && tmp < num){
int x = dfs(tmp, up, down, left, i);
int y = dfs(num - tmp, up, down, i + 1, right);
ans = Min(ans, x + y + down - up + 1);
}
}
return dp[up][down][left][right] = ans;
}
int main(){
int n, m, k;
int kase = 0;
while(scanf("%d%d%d", &n, &m, &k) == 3){
memset(pic, 0, sizeof pic);
memset(dp, -1, sizeof dp);
for(int i = 0; i < k; ++i){
int x, y;
scanf("%d%d", &x, &y);
pic[x][y] = 1;
}
printf("Case %d: %d\n", ++kase, dfs(k, 1, n, 1, m));
}
return 0;
}