{"id":347,"date":"2013-03-08T16:34:00","date_gmt":"2013-03-08T08:34:00","guid":{"rendered":"http:\/\/nike0good.jp1.rpvhost.net\/251"},"modified":"2013-03-08T16:34:00","modified_gmt":"2013-03-08T08:34:00","slug":"poj_1330_nearest_common_ancestors-lca_template_title","status":"publish","type":"post","link":"https:\/\/nike0good.com\/?p=347","title":{"rendered":"POJ 1330(Nearest Common Ancestors-Lca\u6a21\u677f\u9898)"},"content":{"rendered":"\n\n\n
\n
\nNearest Common Ancestors<\/div>\n
\n\n\n\n\n
Time Limit:<\/strong> 1000MS<\/td>\n <\/td>\nMemory Limit:<\/strong> 10000K<\/td>\n<\/tr>\n
Total Submissions:<\/strong> 13559<\/td>\n <\/td>\nAccepted:<\/strong> 7236<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Description<\/p>\n

\nA rooted tree is a well-known data structure in computer science and engineering. An example is shown below: <\/p>\n

\"\" 
\nIn the figure, each node is labeled with an integer from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if node x is in the path between the root and node y. For example, node 4 is an ancestor of node 16. Node 10 is also an ancestor
\n of node 16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16. Remember that a node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the ancestors of node 7. A node x is called a common ancestor of two different nodes y and z if node
\n x is an ancestor of node y and an ancestor of node z. Thus, nodes 8 and 4 are the common ancestors of nodes 16 and 7. A node x is called the nearest common ancestor of nodes y and z if x is a common ancestor of y and z and nearest to y and z among their common
\n ancestors. Hence, the nearest common ancestor of nodes 16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is. <\/p>\n

For other examples, the nearest common ancestor of nodes 2 and 3 is node 10, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common ancestor of nodes 4 and 12 is node 4. In the last example, if y is an ancestor of z, then the nearest
\n common ancestor of y and z is y. <\/p>\n

Write a program that finds the nearest common ancestor of two distinct nodes in a tree. <\/p>\n<\/div>\n

Input<\/p>\n

\nThe input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case starts with a line containing an integer N , the number of nodes in a tree, 2<=N<=10,000. The nodes are labeled with integers 1, 2,...,
\n N. Each of the next N -1 lines contains a pair of integers that represent an edge --the first integer is the parent node of the second integer. Note that a tree with N nodes has exactly N - 1 edges. The last line of each test case contains two distinct integers
\n whose nearest common ancestor is to be computed.<\/div>\n

Output<\/p>\n

\nPrint exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.<\/div>\n

Sample Input<\/p>\n

2\n16\n1 14\n8 5\n10 16\n5 9\n4 6\n8 4\n4 10\n1 13\n6 15\n10 11\n6 7\n10 2\n16 3\n8 1\n16 12\n16 7\n5\n2 3\n3 4\n3 1\n1 5\n3 5\n<\/pre>\n
Sample Output<\/p>\n
4\n3\n\n<\/pre>\n
Source<\/p>\n
\nTaejon 2002<\/a><\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\n<\/span><\/p>\n
Rt,\u7eaf\u7cb9\u65e0\u804a\u5199\u7684\u2026\u2026(\u5176\u5b9e\u81f3\u4eca\u4e3a\u6b62\u5c31\u5199\u8fc72\u4e2a\u2026\u2026\u2026("\u2594\u25a1\u2594)\/)<\/span><\/p>\n
<\/span><\/p>\n
#include<cstdio>\n#include<cstdlib>\n#include<cstring>\n#include<iostream>\n#include<algorithm>\n#include<functional>\nusing namespace std;\n#define MAXN (10000+10)\n#define Li (17)\nint t,n,father[MAXN][Li];\nint edge[MAXN],pre[MAXN],next[MAXN],size=0;\nvoid addedge(int u,int v)\n{\n\tedge[++size]=v;\n\tnext[size]=pre[u];\n\tpre[u]=size;\n}\n\nint queue[MAXN],deep[MAXN],bin[Li];\nvoid bfs()\n{\n\tint head=1,tail=1;\n\tfor (int i=1;i<=n;i++)\n\t\tif (!father[i][0]) {queue[1]=i;deep[i]=1;break;}\n\twhile (head<=tail)\n\t{\n\t\tint &u=queue[head];\n\t\tfor (int i=1;i<Li;i++)\n\t\t{\n\t\t\tfather[u][i]=father[father[u][i-1]][i-1];\n\t\t\tif (father[u][i]==0) break;\n\t\t}\n\t\tfor (int p=pre[u];p;p=next[p])\n\t\t{\n\t\t\tint &v=edge[p];\n\t\t\tdeep[v]=deep[u]+1;\n\t\t\tqueue[++tail]=v;\n\t\t}\n\t\thead++;\n\t}\n}\nint lca(int x,int y)\n{\n\tif (x==y) return x;\n\tif (deep[x]<deep[y]) swap(x,y);\n\tint t=deep[x]-deep[y];\n\tfor (int i=0;i<Li;i++)\n\t\tif (bin[i]&t)\n\t\t{\n\t\t\tx=father[x][i];\n\t\t}\n\tint i=Li-1;\n\twhile (x^y)\n\t{\n\t\twhile (father[x][i]==father[y][i]&&i) i--;\n\t\tx=father[x][i];y=father[y][i];\n\t}\n\treturn x;\n}\nint main()\n{\n\/\/\tfreopen("poj1330.in","r",stdin);\n\tscanf("%d",&t);\n\tfor (int i=0;i<Li;i++) bin[i]=1<<i;\n\n\twhile (scanf("%d",&n)!=EOF)\n\t{\n\t\tmemset(pre,0,sizeof(pre));size=0;\n\t\tmemset(father,0,sizeof(father));\n\t\tfor (int i=1;i<n;i++)\n\t\t{\n\t\t\tint u,v;\n\t\t\tscanf("%d%d",&u,&v);\n\t\t\tfather[v][0]=u;\n\t\t\taddedge(u,v);\n\t\t}\n\t\tbfs();\n\t\tint x,y;\n\t\tscanf("%d%d",&x,&y);\n\t\tprintf("%dn",lca(x,y));\n\n\t}\n\treturn 0;\n}<\/pre>\n<\/p>\n
<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"
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