{
   "question_id": 7420132,
   "question_title": "动态规划-硬币重量最轻问题",
   "question_content": "<p>设有n种不同面值的硬币&#xff0c;第i种硬币的币值是Vi(其中V1&#61;1),重量是Wi,i&#61;1,2&#xff0c;...n且现在购买某种总币值为y的商品&#xff0c;需要用这些硬币付款&#xff0c;如果每种钱币使用的个数不限&#xff0c;那么如何选择付款的方法使得付出钱币的总重量最轻&#xff1f;使用动态规划设计策略设计一个求解该问题的算法。假设问题的输入实例是&#xff1a;</p><p style=\"margin-left:.38in\">V1&#61;1&#xff0c; V2&#61;4&#xff0c; V3&#61;6&#xff0c; V4&#61;8</p><p style=\"margin-left:.38in\">W1&#61;1&#xff0c; W2&#61;2&#xff0c;W3&#61;4&#xff0c;W4&#61;6</p><p style=\"margin-left:.38in\">Y&#61;12</p><p>要求输出优化函数表和标记函数表、以及硬币支付方式。</p>\n",
   "difficulty": "困难",
   "answer_id": 53378192,
   "answer_content": "<pre>\n<code>#include&lt;stdio.h&gt;\nvoid strcpy(int *a, int *b, int Y){\n    for(int i&#61;0;i&lt;&#61;Y;i&#43;&#43;) *(a&#43;i) &#61; *(b&#43;i);\n}\nvoid solve(){\n        int n; scanf(&#34;%d&#34;,&amp;n);\n        int type[n], weight[n], Y, i, j, k;\n        for(i&#61;0;i&lt;n;i&#43;&#43;) scanf(&#34;%d&#34;,&amp;type[i]);\n        for(i&#61;0;i&lt;n;i&#43;&#43;) scanf(&#34;%d&#34;,&amp;weight[i]);\n        scanf(&#34;%d&#34;,&amp;Y);\n        int Min[Y&#43;1], Min_Path[Y&#43;1], path[n][Y&#43;1];\n        for(i&#61;0;i&lt;&#61;Y;i&#43;&#43;) Min[i] &#61; 9999;\n        Min[0] &#61; 0;\n        printf(&#34;\\n&#34;);\n        for(j&#61;0;j&lt;n;j&#43;&#43;){\n            for(i&#61;type[j]; i&lt;&#61;Y; i&#43;&#43;)\n                if(Min[i] &gt; Min[i-type[j]]&#43;weight[j]){\n                    Min_Path[i] &#61; type[j];\n                    Min[i] &#61; Min[i-type[j]]&#43;weight[j];\n                }\n            for(k&#61;1;k&lt;&#61;Y;k&#43;&#43;) printf(&#34;%-3d&#34;,Min[k]);\n            printf(&#34;\\n&#34;);\n            strcpy(path[j],Min_Path,Y);\n        }\n\n        printf(&#34;\\n&#34;);\n        for(i&#61;0;i&lt;n;i&#43;&#43;){\n            for(j&#61;1;j&lt;&#61;Y;j&#43;&#43;)\n                printf(&#34;%-3d&#34;,path[i][j]);\n            printf(&#34;\\n&#34;);\n        }\n\n        int y&#61;Y;\n        printf(&#34;\\n支付方式:&#34;);\n        while (y){\n            printf(&#34;%d &#34;,Min_Path[y]);\n            y -&#61; Min_Path[y];\n        }\n        printf(&#34;\\n总重量:%d\\n&#34;,Min[Y]);\n}\nint main(){\n    solve();\n    return 1;\n}\n/*\n4\n1 4 6 8\n1 3 2 6\n12\n */</code></pre>\n\n<p style=\"text-align:center\"><img alt=\"\" height=\"492\" src=\"https://img-ask.csdnimg.cn/upload/1619149096414.png\" width=\"449\" /></p>\n\n<p> </p>",
   "tag_name": "c语言",
   "cpp": "#include<stdio.h>\nvoid strcpy(int *a, int *b, int Y){\n\tfor(int i=0;i<=Y;i++) *(a+i) = *(b+i);\n}\nvoid solve(){\n\tint n; scanf(\"%d\",&n);\n\tint type[n], weight[n], Y, i, j, k;\n\tfor(i=0;i<n;i++) scanf(\"%d\",&type[i]);\n\tfor(i=0;i<n;i++) scanf(\"%d\",&weight[i]);\n\tscanf(\"%d\",&Y);\n\tint Min[Y+1], Min_Path[Y+1], path[n][Y+1];\n\tfor(i=0;i<=Y;i++) Min[i] = 9999;\n\tMin[0] = 0;\n\tprintf(\"\\n\");\n\tfor(j=0;j<n;j++){\n\t\tfor(i=type[j]; i<=Y; i++)\n\t\t\tif(Min[i] > Min[i-type[j]]+weight[j]){\n\t\t\t\tMin_Path[i] = type[j];\n\t\t\t\tMin[i] = Min[i-type[j]]+weight[j];\n\t\t\t}\n\t\tfor(k=1;k<=Y;k++) printf(\"%-3d\",Min[k]);\n\t\tprintf(\"\\n\");\n\t\tstrcpy(path[j],Min_Path,Y);\n\t}\n\tprintf(\"\\n\");\n\tfor(i=0;i<n;i++){\n\t\tfor(j=1;j<=Y;j++)\n\t\t\tprintf(\"%-3d\",path[i][j]);\n\t\tprintf(\"\\n\");\n\t}\n\tint y=Y;\n\tprintf(\"\\n支付方式:\");\n\twhile (y){\n\t\tprintf(\"%d \",Min_Path[y]);\n\t\ty -= Min_Path[y];\n\t}\n\tprintf(\"\\n总重量:%d\\n\",Min[Y]);\n}\nint main(){\n\tsolve();\n\treturn 1;\n}",
   "topic_link": "https://bbs.csdn.net/topics/600470163",
   "status": 1,
   "keywords": "算法中阶,动态规划,动态规划原理,高级设计和分析技术",
   "license": "csdn.net",
   "notebook": {
      "cpp": "https://codechina.csdn.net/csdn/csdn-daily-code/-/jupyter/master/data/notebook/answer/ipynb/cpp/70.ipynb?type=file"
   },
   "notebook_enable": 1
}