{"id":10211,"date":"2025-03-16T07:23:47","date_gmt":"2025-03-16T14:23:47","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=10211"},"modified":"2025-03-16T08:25:38","modified_gmt":"2025-03-16T15:25:38","slug":"leetcode-3489-zero-array-transformation-iv","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/dynamic-programming\/leetcode-3489-zero-array-transformation-iv\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 3489. Zero Array Transformation IV"},"content":{"rendered":"\n<p>\u4e00\u9053\u4e0d\u9519\u7684DP\u9898\u76ee\uff01<\/p>\n\n\n\n<p>\u9996\u5148\u770b\u5230\u6bcf\u6b21\u53ef\u4ee5\u53d6\u4efb\u610f\u4e00\u4e2anums[l] ~ nums[r]\u7684\u5b50\u96c6 <\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>\u4e0d\u80fd\u7528\u8d2a\u5fc3\uff08\u65e0\u6cd5\u627e\u5230\u5168\u5c40\u6700\u4f18\u89e3\uff09<\/li><li>\u4e0d\u80fd\u7528\u641c\u7d22 \uff08\u6570\u636e\u89c4\u6a21\u592a\u5927\uff0c(2^10) ^ 1000\uff09<\/li><\/ul>\n\n\n\n<p>\u90a3\u53ea\u80fd\u7528\u52a8\u6001\u89c4\u5212\u4e86<\/p>\n\n\n\n<p>\u72b6\u6001\u5b9a\u4e49: dp[k][i][j] \u80fd\u5426\u901a\u8fc7\u4f7f\u7528\u524dk\u4e2a\u53d8\u6362\u4f7f\u5f97\u7b2ci\u4e2a\u6570\u7684\u503c\u53d8\u6210j<\/p>\n\n\n\n<p>\u8fb9\u754c\u6761\u4ef6: dp[0][i][nums[i]] = 1\uff0c\u4e0d\u4f7f\u7528\u4efb\u4f55\u53d8\u6362\uff0c\u7b2ci\u4e2a\u6570\u53ef\u4ee5\u8fbe\u5230\u7684\u6570\u503c\u5c31\u662fnums[i]\u672c\u8eab\u3002<\/p>\n\n\n\n<p>\u72b6\u6001\u8f6c\u79fb\uff1adp[k][i][j] = dp[k-1][i][j] | (dp[k &#8211; 1][i][j + val[k]] if l[k] &lt;= i &lt;= r[k] else 0)<\/p>\n\n\n\n<p>\u7b80\u5355\u6765\u8bf4\u5982\u679c\u7b2ck-1\u8f6e\u7b2ci\u4e2a\u6570\u53ef\u4ee5\u53d8\u6210j + val[k]\uff0c\u90a3\u4e48\u7b2ck\u8f6e\u5c31\u53ef\u4ee5\u901a\u8fc7\u51cf\u53bbval[k]\u53d8\u6210j\u3002<\/p>\n\n\n\n<p>\u4e0a\u9762\u662f\u62c9\u7684\u516c\u5f0f\uff0c\u6211\u4eec\u4e5f\u53ef\u4ee5\u5199\u6210\u63a8\u7684:<\/p>\n\n\n\n<p>dp[k][i][j &#8211; val[k]] = dp[k-1][i][j &#8211; vak[k]] | (dp[k &#8211; 1][i][j] if l[k] &lt;= i &lt;= r[k] else 0)<\/p>\n\n\n\n<p>\u5f53\u7136\u8fd9\u4e48\u5b9a\u4e49\u7684\u8bdd\u7a7a\u95f4\u590d\u6742\u5ea6\u592a\u9ad8O(10^7)\uff0c\u7531\u4e8e\u7b2ck\u8f6e\u7684\u521d\u59cb\u72b6\u6001\u5c31\u7b49\u4e8ek-1\u8f6e\u7684\u72b6\u6001\uff0c\u6211\u4eec\u53ef\u4ee5\u4f7f\u7528\u6eda\u52a8\u6570\u7ec4\u6765\u964d\u7ef4\uff0c\u7a7a\u95f4\u590d\u6742\u5ea6\u964d\u4f4e\u5230O(10^4)\u3002<\/p>\n\n\n\n<p>\u65f6\u95f4\u590d\u6742\u5ea6\uff1aO(k*n*MaxV) = O(10^7)\u3002<\/p>\n\n\n\n<pre lang=\"c++\">\nclass Solution {\npublic:\n    int minZeroArray(vector<int>& nums, vector<vector<int>>& queries) {\n      if (accumulate(begin(nums), end(nums), 0) == 0) return 0;\n      constexpr int kMaxV = 1000;\n      const int n = nums.size();\n      const int K = queries.size();\n      vector<bitset<kMaxV + 1>> dp(n);\n      for (int i = 0; i < n; ++i)\n        dp[i][nums[i]] = 1;\n      for (int k = 0; k < K; ++k) {\n        int l = queries[k][0];\n        int r = queries[k][1];\n        int v = queries[k][2];\n        for (int i = l; i <= r; ++i)\n          for (int j = 0; j <= kMaxV - v; ++j)\n            if (dp[i][j + v]) dp[i][j] = 1;\n        bool found = true;\n        for (int i = 0; i < n; ++i)\n          if (!dp[i][0]) found = false;\n        if (found) return k + 1;\n      }\n      return -1;\n    }\n};\n<\/pre>\n\n\n\n<p>\u526a\u679d\u4f18\u5316<\/p>\n\n\n\n<p>\u4e0a\u9762\u6211\u4eec\u628a\u6574\u4e2a\u6570\u7ec4\u770b\u4f5c\u4e00\u4e2a\u6574\u4f53\uff0c\u4e00\u8f6e\u4e00\u8f6e\u6765\u505a\u3002\u4f46\u5982\u679c\u67d0\u4e9b\u6570\u5728\u7b2ck\u8f6e\u80fd\u5230\u53d8\u62100\u4e86\uff0c\u5c31\u6ca1\u6709\u5fc5\u8981\u53c2\u4e0e\u540e\u9762\u7684\u53d8\u5316\u4e86\uff0c\u6216\u8005\u8bf4\u5b83\u7528\u4e8e\u65e0\u6cd5\u53d8\u62100\uff0c\u90a3\u4e48\u5c31\u662f\u65e0\u89e3\uff0c\u5176\u4ed6\u6570\u4e5f\u5c31\u4e0d\u9700\u8981\u518d\u8ba1\u7b97\u4e86\u3002<\/p>\n\n\n\n<p>\u6240\u4ee5\uff0c\u6211\u4eec\u53ef\u4ee5\u5bf9<strong><em>\u6bcf\u4e2a\u6570\u5355\u72ec\u8fdb\u884cdp<\/em><\/strong>\u3002\u5373\u5bf9\u4e8e\u7b2ci\u4e2a\u6570\uff0c\u8ba1\u7b97\u6700\u5c11\u9700\u8981\u591a\u5c11\u8f6e\u624d\u80fd\u628a\u5b83\u53d8\u62100\u3002\u7136\u540e\u5bf9\u6240\u6709\u7684\u8f6e\u6570\u53d6\u4e00\u4e2a\u6700\u5927\u503c\u3002\u603b\u7684\u65f6\u95f4\u590d\u6742\u5ea6\u4e0d\u53d8\uff08\u6700\u574f\u60c5\u51b5\u6240\u6709\u6570\u90fd\u9700\u8981\u7ecf\u8fc7K\u8f6e\uff09\u3002\u7a7a\u95f4\u590d\u6742\u5ea6\u5219\u53ef\u4ee5\u518d\u964d\u4f4e\u4e00\u4e2a\u7eac\u5ea6\u5230O(MaxV) = O(10^3)\u3002<\/p>\n\n\n\n<pre lang=\"c++\">\nclass Solution {\npublic:\n    int minZeroArray(vector<int>& nums, vector<vector<int>>& queries) {\n      constexpr int kMaxV = 1000;\n      const int n = nums.size();\n      const int K = queries.size();\n      auto rounds = [&](int i) -> int {\n        if (!nums[i]) return 0;\n        bitset<kMaxV + 1> dp;\n        dp[nums[i]] = 1;\n        for (int k = 0; k < K; ++k) {\n          int l = queries[k][0];\n          int r = queries[k][1];\n          int v = queries[k][2];\n          if (l > i || r < i) continue;\n          for (int j = 0; j <= kMaxV - v; ++j)\n            if (dp[j + v]) dp[j] = 1;\n          if (dp[0]) return k + 1;\n        }\n        return INT_MAX;\n      };\n      int ans = 0;\n      for (int i = 0; i < n &#038;&#038; ans <= K; ++i)\n        ans = max(ans, rounds(i));\n      return ans == INT_MAX ? -1 : ans;\n    }\n};\n<\/pre>\n\n\n\n<p>\u518d\u52a0\u901f\uff1a\u6211\u4eec\u53ef\u4ee5\u4f7f\u7528C++ bitset\u7684\u53f3\u79fb\u64cd\u4f5c\u7b26\uff0cdp &gt;&gt; v\uff0c\u76f8\u5f53\u4e8e\u628a\u6574\u4e2a\u96c6\u5408\u5168\u90e8\u526a\u53bbv\uff0c\u518d\u548c\u539f\u5148\u7684\u72b6\u6001\u505a\u6216\u8fd0\u7b97\uff08\u96c6\u5408\u5e76\uff09\u6765\u8fbe\u5230\u65b0\u7684\u72b6\u6001\u3002\u65f6\u95f4\u590d\u6742\u5ea6\u5e94\u8be5\u662f\u4e00\u6837\u7684\uff0c\u53ea\u662f\u4ee3\u7801\u7b80\u5355\uff0c\u901f\u5ea6\u4e5f\u4f1a\u5feb\u4e0d\u5c11\u3002\u6ce8: dp&gt;&gt;v \u4f1a\u521b\u5efa\u4e00\u4e2a\u4e34\u65f6\u5bf9\u8c61\uff0c\u5927\u5c0f\u4e3aO(MaxV)\u3002<\/p>\n\n\n\n<p>\u4e3e\u4e2a\u4f8b\u5b50\uff1a<br>dp = {2, 3, 7} <br>dp &gt;&gt; 2 -&gt; {0, 1, 5}<br>dp |= dp &gt;&gt; v -&gt; {0, 1, 2, 3, 5, 7]<\/p>\n\n\n\n<pre lang=\"c++\">\nclass Solution {\npublic:\n    int minZeroArray(vector<int>& nums, vector<vector<int>>& queries) {\n      constexpr int kMaxV = 1000;\n      const int n = nums.size();\n      const int K = queries.size();\n      auto rounds = [&](int i) -> int {\n        if (!nums[i]) return 0;\n        bitset<kMaxV + 1> dp;\n        dp[nums[i]] = 1;\n        for (int k = 0; k < K; ++k) {\n          int l = queries[k][0];\n          int r = queries[k][1];\n          int v = queries[k][2];\n          if (l > i || r < i) continue;\n          dp |= dp >> v; \/\/ magic\n          if (dp[0]) return k + 1;\n        }\n        return INT_MAX;\n      };\n      int ans = 0;\n      for (int i = 0; i < n &#038;&#038; ans <= K; ++i)\n        ans = max(ans, rounds(i));\n      return ans == INT_MAX ? -1 : ans;\n    }\n};\n<\/pre>\n","protected":false},"excerpt":{"rendered":"<p>\u4e00\u9053\u4e0d\u9519\u7684DP\u9898\u76ee\uff01 \u9996\u5148\u770b\u5230\u6bcf\u6b21\u53ef\u4ee5\u53d6\u4efb\u610f\u4e00\u4e2anums[l] ~ nums[r]\u7684\u5b50\u96c6 \u4e0d\u80fd\u7528\u8d2a\u5fc3\uff08\u65e0\u6cd5\u627e\u5230\u5168\u5c40\u6700\u4f18\u89e3\uff09 \u4e0d\u80fd\u7528\u641c\u7d22 \uff08\u6570\u636e\u89c4\u6a21\u592a\u5927\uff0c(2^10) ^ 1000\uff09 \u90a3\u53ea\u80fd\u7528\u52a8\u6001\u89c4\u5212\u4e86 \u72b6\u6001\u5b9a\u4e49: dp[k][i][j] \u80fd\u5426\u901a\u8fc7\u4f7f\u7528\u524dk\u4e2a\u53d8\u6362\u4f7f\u5f97\u7b2ci\u4e2a\u6570\u7684\u503c\u53d8\u6210j \u8fb9\u754c\u6761\u4ef6: dp[0][i][nums[i]] = 1\uff0c\u4e0d\u4f7f\u7528\u4efb\u4f55\u53d8\u6362\uff0c\u7b2ci\u4e2a\u6570\u53ef\u4ee5\u8fbe\u5230\u7684\u6570\u503c\u5c31\u662fnums[i]\u672c\u8eab\u3002 \u72b6\u6001\u8f6c\u79fb\uff1adp[k][i][j] = dp[k-1][i][j] | (dp[k &#8211; 1][i][j +&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[46],"tags":[18,6,177],"class_list":["post-10211","post","type-post","status-publish","format-standard","hentry","category-dynamic-programming","tag-dp","tag-leetcode","tag-medium","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/10211","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=10211"}],"version-history":[{"count":10,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/10211\/revisions"}],"predecessor-version":[{"id":10227,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/10211\/revisions\/10227"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=10211"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=10211"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=10211"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}