uva1629

題目描述：有一個n行m列（1≤n，m≤20）的網格蛋糕上有一些櫻桃。每次可以用一刀沿着網格線把 蛋糕切成兩塊，並且只能夠直切不能拐彎。要求最後每一塊蛋糕上恰好有一個櫻桃，且切割 線總長度最小。

分析：dp切割行和列

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<queue>
#include<map>
#include<sstream>
#include<cassert>
#include<set>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int maxn = 25;
int n, m;
int dp[maxn][maxn][maxn][maxn];
int g[maxn][maxn];
int judge(int u, int d, int l, int r) {
	int ok = 0;
	for (int i = d; i <= u; i++) {
		for (int j = l; j <= r; j++) {
			if (g[i][j]) {
				if (ok)return 2;
				else ok = 1;
			}
		}
	}
	return ok;
}

int solve(int u, int d, int l, int r) {
	int &k = dp[u][d][l][r];
	if (k != -1)return k;
	int t = judge(u, d, l, r);
	if (t == 1)return 0;
	else if (t == 0)return INF;
	k = INF;
	for (int i = l; i < r; i++) {
		k = min(k, solve(u, d, l, i) + solve(u, d, i + 1, r) + u - d + 1);
	}
	for (int i = d; i < u; i++) {
		k = min(k, solve(i, d, l, r) + solve(u, i+1, l, r) + r - l + 1);
	}
	return k;
}
int main() {
	int k, t = 0;
	while (cin >> n >> m >> k) {
		memset(g, 0, sizeof(g));
		memset(dp, -1, sizeof(dp));
		int u, v;
		while (k--) {
			cin >> u >> v;
			g[u][v] = 1;
		}
		cout << "Case " << ++t << ": " << solve(n, 1, 1, m) << endl;;
	}
	return 0;
}