UVa 1629 Cake slicing (记忆化搜索)

时间：2021-02-27 10:52:21

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;

}

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