代码随想录算法训练营第四十六天|139.单词拆分 多重背包问题

目录

LeeCode 139.单词拆分

多重背包问题

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LeeCode 139.单词拆分

139. 单词拆分 - 力扣（LeetCode）

动归五部曲：

1.确定dp数组及下标含义: dp[i]: 字符串长度为i的话，dp[i]为true，表示可以拆分为一个或多个在字典中出现的单词；

2.确定递推公式： if([j, i] 这个区间的子串出现在字典里 && dp[j]是true) {dp[i] = true； }

3.dp数组如何初始化：dp[0] = true; dp[i] = false;

4.确定遍历顺序：先遍历 背包，再遍历物品;

5.举例递推dp数组

代码：

class Solution {
public:
    bool wordBreak(string s, vector& wordDict) {
    	unordered_set wordSet(wordDict.begin(), wordDict.end());
		vector dp(s.size() + 1, false);
		dp[0] = true;
		for (int i = 1; i <= s.size(); i++) {
			for (int j = 0; j < i; j++) {
				string word = s.substr(j, i - j);
				if (wordSet.find(word) != wordSet.end() && dp[j]) {
					dp[i] = true;
				}
			}
		} 
        return dp[s.size()];
    }
};

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多重背包问题

每件物品最多有Mi件可用，把Mi件摊开，可以转化成01背包问题。

代码：

版本1：

void test_multi_pack() {
	vector weight = {1, 3, 4};
	vector value = {15, 20, 30};
	vector nums = {2, 3, 2};
	int bagWeight = 10;
	for (int i = 0; i < nums.size(); i++) {
		while (nums[i] > 1) {
			weight.push_back(weight[i]);
			value.push_back(value[i]);
			nums[i]--;
		}
	}
	vector dp(bagWeight + 1, 0);
	for (int i = 0; i < weight.size(); i++) {
		for (int j = bagWeight; j >= weight[i]; j--) {
			dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
		}
		for (int j = 0; j <= bagWeight; j++) {
			cout << dp[j] << " ";
		}
		cout << endl;
	}
	cout << dp[bagWeight] << endl;
}
int main() {
	test_multi_pack();
}

版本2：

void test_multi_pack() {
	vector weight = {1, 3, 4};
	vector value = {15, 20, 30};
	vector nums = {2, 3, 2};
	int bagWeight = 10;
	vector dp(bagWeight + 1, 0);
	for (int i = 0; i < weight.size(); i++) {
		for (int j = bagWeight; j >= weight[i]; j--) {
			for (int k = 1; k <= nums[i] && (j - k * weight[i]) >= 0; k++) {
				dp[j] = max(dp[j], dp[j - k * weight[i]] + k * value[i]);
			}
		}
		for (int j = 0; j <= bagWeight; j++) {
			cout << dp[j] << " ";
		}
		cout << endl;
	}
	cout << dp[bagWeight] << endl;
}
int main() {
	test_multi_pack();
}