目录
LeeCode 139.单词拆分
多重背包问题
LeeCode 139.单词拆分
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<string>& wordDict) {
unordered_set<string> wordSet(wordDict.begin(), wordDict.end());
vector<bool> 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()];
}
};
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<int> weight = {1, 3, 4};
vector<int> value = {15, 20, 30};
vector<int> 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<int> 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<int> weight = {1, 3, 4};
vector<int> value = {15, 20, 30};
vector<int> nums = {2, 3, 2};
int bagWeight = 10;
vector<int> 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();
}