uva 10118 free candies

来源：互联网发布：中国有多少穷人知乎编辑：程序博客网时间：2024/05/29 07:05

题目大意：有4堆糖果，每堆糖果有n颗糖果，然后给出每颗糖果的类型1~20，然后只有取走当前堆的前面一颗糖果后才可以取后面的糖果，然后小伙伴有一个篮子，篮子可以装5个糖果，如果篮子中的糖果存在相同类型的两个糖果，便可以将这两颗糖果算成一对放进腰包，问，小伙伴按什么样的方式取糖果可以取到最多的糖果。

解题思路：记忆化搜索，开一个四维数组记录当前4堆糖果取走相应个数后能拿到的最多对糖果数。

//  Created by Chenhongwei in 2015.//  Copyright (c) 2015 Chenhongwei. All rights reserved.#include"iostream"#include"cstdio"#include"cstdlib"#include"cstring"#include"climits"#include"queue"#include"cmath"#include"map"#include"set"#include"stack"#include"vector"#include"sstream"#include"algorithm"using namespace std;typedef long long ll;int g[50][5], dp[50][50][50][50], vis[25], b[10];int n;int dfs(int cnt){int &cur = dp[b[0]][b[1]][b[2]][b[3]];if (cnt == 5)return cur = 0;if (cur != -1)return cur;cur = 0;for (int i = 0; i < 4; i++){if (b[i] == n)continue;int tmp = g[b[i]++][i];if (vis[tmp]){vis[tmp] = 0;cur = max(cur, dfs(cnt - 1) + 1);vis[tmp] = 1;}else{vis[tmp] = 1;cur = max(cur, dfs(cnt + 1));vis[tmp] = 0;b[i]--;}return cur;}int main(){//ios::sync_with_stdio(false);// freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);while (cin >> n && n){memset(dp, -1, sizeof dp);memset(vis, 0, sizeof vis);memset(b, 0, sizeof b);for (int i = 0; i < n; i++)for (int j = 0; j < 4; j++)cin >> g[i][j];cout << dfs(0) << endl;}return 0;}

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