uva 1629切蛋糕（dp）

有一个n行m列的网格蛋糕，上面有一些樱桃。求使得每块蛋糕上都有一个樱桃的分割最小长度

思路：dp。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#define eps 1e-6
#define LL long long
using namespace std;  

const int maxn = 100 + 5;
const int INF = 10000000;
int d[22][22][22][22];
int n, m, K, kase = 0;//樱桃数量
int cherry[22][22], sumv[22][22];

int cpu(int a, int b, int c, int d) {
	return sumv[b][d] - sumv[a-1][d] - sumv[b][c-1] + sumv[a-1][c-1];
}

int dp(int begx, int endx, int begy, int endy, int cherrynum) {
	int& ans = d[begx][endx][begy][endy];
	if(cherrynum == 1) return 0;
	if(ans != -1) return ans;
	ans = INF;
	for(int i = begx; i < endx; i++) if(cpu(begx, i, begy, endy) > 0 && cpu(begx, i, begy, endy) < cherrynum)
		ans = min(ans, dp(begx, i, begy, endy, cpu(begx, i, begy, endy))+dp(i+1, endx, begy, endy, cherrynum-cpu(begx, i, begy, endy))+endy-begy+1);
	for(int i = begy; i < endy; i++) if(cpu(begx, endx, begy, i) > 0 && cpu(begx, endx, begy, i) < cherrynum)
		ans = min(ans, dp(begx, endx, begy, i, cpu(begx, endx, begy, i))+dp(begx, endx, i+1, endy, cherrynum-cpu(begx, endx, begy, i))+endx-begx+1);
	return ans;
}

void init() {
	memset(d, -1, sizeof(d));
	memset(cherry, 0, sizeof(cherry));
	memset(sumv, 0, sizeof(sumv));
	int x, y;
	for(int i = 0; i < K; i++) {
		cin >> x >> y;
		cherry[x][y] = 1;
	}
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++)
			sumv[i][j] = sumv[i][j-1] + sumv[i-1][j] + cherry[i][j] - sumv[i-1][j-1];
}

void solve() {
	printf("Case %d: %d\n", ++kase, dp(1, n, 1, m, K));
}

int main() {
	//freopen("input.txt", "r", stdin);
	while(scanf("%d%d%d", &n, &m, &K) == 3) {
		init();
		solve();
	}
	return 0;
}