畅通工程最小生成树普利姆算法和克鲁斯卡尔算法解题浙大计算机研究生复试上机考试-2007年

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 <td> <h2>畅通工程</h2> Time Limit: 1000/1000 MS (Java/Others)&nbsp;&nbsp;&nbsp;&nbsp;Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 38345&nbsp;&nbsp;&nbsp;&nbsp;Accepted Submission(s): 17093 &nbsp; Problem Description 省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可）。经过调查评估，得到的统计表中列出了有可能建设公路的若干条道路的成本。现请你编写程序，计算出全省畅通需要的最低成本。 &nbsp; &nbsp; Input 测试输入包含若干测试用例。每个测试用例的第1行给出评估的道路条数 N、村庄数目M ( &lt; 100 )；随后的 N 行对应村庄间道路的成本，每行给出一对正整数，分别是两个村庄的编号，以及此两村庄间道路的成本（也是正整数）。为简单起见，村庄从1到M编号。当N为0时，全部输入结束，相应的结果不要输出。 &nbsp; &nbsp; Output 对每个测试用例，在1行里输出全省畅通需要的最低成本。若统计数据不足以保证畅通，则输出“?”。 &nbsp; &nbsp; Sample Input <pre><code>&nbsp;</code></pre> 3 3 1 2 1 1 3 2 2 3 4 1 3 2 3 2 0 100 &nbsp; &nbsp; Sample Output <pre><code>&nbsp;</code></pre> 3 ? &nbsp; &nbsp; Source <a href="http://acm.hdu.edu.cn/search.php?field=problem&amp;key=%D5%E3%B4%F3%BC%C6%CB%E3%BB%FA%D1%D0%BE%BF%C9%FA%B8%B4%CA%D4%C9%CF%BB%FA%BF%BC%CA%D4-2007%C4%EA&amp;source=1&amp;searchmode=source" rel="nofollow">浙大计算机研究生复试上机考试-2007年 </a> &nbsp; &nbsp; Recommend lcy&nbsp;&nbsp;&nbsp;|&nbsp;&nbsp;&nbsp;We have carefully selected several similar problems for you:&nbsp;&nbsp;<a href="http://acm.hdu.edu.cn/showproblem.php?pid=1879" rel="nofollow">1879</a>&nbsp;<a href="http://acm.hdu.edu.cn/showproblem.php?pid=1233" rel="nofollow">1233</a>&nbsp;<a href="http://acm.hdu.edu.cn/showproblem.php?pid=1875" rel="nofollow">1875</a>&nbsp;<a href="http://acm.hdu.edu.cn/showproblem.php?pid=1232" rel="nofollow">1232</a>&nbsp;<a href="http://acm.hdu.edu.cn/showproblem.php?pid=1102" rel="nofollow">1102</a>&nbsp; &nbsp; 今天是图论集训第一天，以前讲数据结构图的时候，对图怕怕的，但怕又能怎样？只有坦然面对，有意思的东西固然难，加油，蓝孩，为了你的Jeep，加油！ &nbsp; #include&lt;iostream&gt; #include&lt;algorithm&gt; #include&lt;string.h&gt; using namespace std; struct kk { &nbsp;&nbsp; &nbsp;int s,e,w; }; int Prim(int N,int M) &nbsp; { &nbsp;&nbsp; &nbsp;kk a[10000]; &nbsp;&nbsp; &nbsp;int i,j,pos=1,min,flag=0,result=0,map[101][101],low[101],visit[101]; &nbsp; &nbsp; low数组存当前到各个顶点的最小权值 &nbsp;&nbsp; &nbsp;memset(map,0x3f3f3f3f,sizeof(map)); &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 将邻接矩阵设为无穷大 &nbsp;&nbsp; &nbsp;memset(visit,0,sizeof(visit)); &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;for(i=0;i&lt;N;i++) &nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;scanf("%d%d%d",&amp;a[i].s,&amp;a[i].e,&amp;a[i].w); &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;map[a[i].s][a[i].e]=a[i].w; &nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;map[a[i].e][a[i].s]=a[i].w; &nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;visit[pos]=1; &nbsp;&nbsp; &nbsp;for(i=1;i&lt;=M;i++) &nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(!visit[i]) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;low[i]=map[pos][i]; &nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;for(i=0;i&lt;M-1;i++) &nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;min=0x3f3f3f3f; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;for(j=1;j&lt;=M;j++) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(!visit[j]&amp;&amp;low[j]&lt;min) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;min=low[j]; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;pos=j; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;result+=min; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;visit[pos]=1; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;for(j=1;j&lt;=M;j++) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(!visit[j]&amp;&amp;low[j]&gt;map[pos][j]) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;low[j]=map[pos][j]; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; &nbsp;for(i=1;i&lt;=M;i++) &nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(visit[i]==0) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;flag=1; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;break; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;if(flag==1) &nbsp;&nbsp; &nbsp;return 0; &nbsp;&nbsp; &nbsp;else &nbsp;&nbsp; &nbsp;return result; } int main() { &nbsp;&nbsp; &nbsp;int n,m; &nbsp;&nbsp; &nbsp; &nbsp;&nbsp; &nbsp;while(~scanf("%d%d",&amp;n,&amp;m)) &nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(n==0) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;break; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;int t=Prim(n,m); &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(t==0) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;cout&lt;&lt;"?"&lt;&lt;endl; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;else &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;cout&lt;&lt;t&lt;&lt;endl; &nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;return 0; } &nbsp; 下面是克鲁斯卡尔算法 #include&lt;iostream&gt; #include&lt;string&gt; #include&lt;algorithm&gt; using namespace std; struct kk { &nbsp; &nbsp;int s,e,w; }s[10009]; int a[105]; int cmp(kk x,kk y) { &nbsp;&nbsp; &nbsp;return x.w&lt;y.w; } int find(int x) { &nbsp;&nbsp; &nbsp;return x==a[x]?x:a[x]=find(a[x]); } int join(int x,int y) { &nbsp;&nbsp; &nbsp;int fx=find(x); &nbsp;&nbsp; &nbsp;int fy=find(y); &nbsp;&nbsp; &nbsp;if(fx!=fy) &nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;a[fx]=fy; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;return 1; &nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;else &nbsp;&nbsp; &nbsp;return 0; } int main() { &nbsp;&nbsp; &nbsp;int i,j,n,m; &nbsp;&nbsp; &nbsp;//freopen("1.txt","r",stdin); &nbsp;&nbsp; &nbsp;while(~scanf("%d%d",&amp;n,&amp;m)) &nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(n==0) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;break; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;int f=0,sum=0; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;for(i=1;i&lt;=n;i++) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;cin&gt;&gt;s[i].s&gt;&gt;s[i].e&gt;&gt;s[i].w; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;for(i=1;i&lt;=m;i++) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;a[i]=i;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;sort(s+1,s+n+1,cmp); &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;for(i=1;i&lt;=n;i++) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(join(s[i].s,s[i].e)) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;{ &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;f++; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;sum+=s[i].w; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(f==m-1) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;break; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(f==m-1) &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;cout&lt;&lt;sum&lt;&lt;endl; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;else &nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;cout&lt;&lt;"?"&lt;&lt;endl; &nbsp;&nbsp; &nbsp;} &nbsp;&nbsp; &nbsp;return 0; } &nbsp; </td> 
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