标签：691 target int Stickers sticker Spell res stickers dp
问题：
给定一堆单词卡，每个单词可以有无限张同样的卡片。
要从这些单词卡中选择一些，进行剪切拼接，要构成目标单词，求需要的最少卡片数量。
Example 1: Input: ["with", "example", "science"], "thehat" Output: 3 Explanation: We can use 2 "with" stickers, and 1 "example" sticker. After cutting and rearrange the letters of those stickers, we can form the target "thehat". Also, this is the minimum number of stickers necessary to form the target string. Example 2: Input: ["notice", "possible"], "basicbasic" Output: -1 Explanation: We can't form the target "basicbasic" from cutting letters from the given stickers. Note: stickers has length in the range [1, 50]. stickers consists of lowercase English words (without apostrophes). target has length in the range [1, 15], and consists of lowercase English letters. In all test cases, all words were chosen randomly from the 1000 most common US English words, and the target was chosen as a concatenation of two random words. The time limit may be more challenging than usual. It is expected that a 50 sticker test case can be solved within 35ms on average.
解法：DFS（深度优先搜索）
- 当前状态：
- 当前目标串target，所需要的最少卡片数。
- 选择：
- 在所有单词卡中选择，选择后，剩下的字符再进行递归匹配，得到最少卡片数+1，为当前选择的结果。
- 在这些结果的有效结果（剩下字符递归结果!=-1）中，求res=最小值min
- 将最小值最为本次递归的结果返回 return res，同时保存在结果map：dp中：dp[target]=res。
- 退出递归条件：
- dp中保存过，当前目标串target的结果，直接返回dp[target]
- 当前target串=""，则返回0。（或者提前保存dp[""]=0）
♻️ 优化：
Optimization: If the target can be spelled out by a group of stickers, at least one of them has to contain character target[0]. So I explicitly require next sticker containing target[0], which significantly reduced the search space.
// 这个优化剪枝很神奇，可以避免潜在的对target完全没有贡献的sticker，暂时无法贡献的sticker等到其他sticker把部分target拼凑出来以后再贡献
代码参考：
1 class Solution {
2 public:
3 //dp[s]: for string s, the min number of stickers need.
4 unordered_map<string, int> dp;
5 int dfs(vector<vector<int>>mp, string target) {
6 //return already gotten result.
7 if(dp.count(target)) return dp[target];
8 //if not already got, then following process:
9 int res = INT_MAX;
10 //get alphbet map of this target string.
11 vector<int>tgmp(26,0);
12 for(char a:target) {
13 tgmp[a-'a']++;
14 }
15 //opt: check every sticker
16 for(int j=0; j<mp.size(); j++) {
17 //every alphbet of this sticker
18 //optimization:
19 //If the target can be spelled out by a group of stickers,
20 //at least one of them has to contain character target[0].
21 //So I explicitly require next sticker containing target[0],
22 //which significantly reduced the search space.
23 if(mp[j][target[0]-'a']==0)continue;
24 //process
25 string nexttarget;
26 for(int i=0; i<26; i++) {
27 if(tgmp[i]>mp[j][i]) {
28 int leftAlphbetCount = tgmp[i]-mp[j][i];
29 nexttarget += string(leftAlphbetCount, 'a'+i);
30 }
31 }
32 // if (nexttarget == target) continue;
33 // 如果没有之前那个优化，一定要判断新target和原target是否一样，避免重复递归
34 int tmp = dfs(mp, nexttarget);
35 if(tmp!=-1) res = min(res, tmp+1);
36 }
37 dp[target] = (res==INT_MAX?-1:res);
38 return dp[target];
39 }
40 int minStickers(vector<string>& stickers, string target) {
41 //get alphbet map of every word
42 int sn = stickers.size();
43 vector<vector<int>>mp(sn, vector<int>(26, 0));
44 for(int i=0; i<sn; i++) {
45 for(char a:stickers[i]) {
46 mp[i][a-'a']++;
47 }
48 }
49 dp[""]=0;
50 return dfs(mp, target);
51 }
52 };
标签：691,target,int,Stickers,sticker,Spell,res,stickers,dp 来源： https://www.cnblogs.com/habibah-chang/p/14321782.html
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