Sicily 1140 国王的遗产

N个黄金N-1条链，是树结构。

盲搜，搜索状态空间：每次切掉一条边后出现的两个子图。问题转化为怎么样表示和比较这些子图。

显而易见，其中一个子图是原树的一个子树，另一个子图就是该子树的补图。子树可以用它的根来表示。再考虑如何比较它们的优劣，如果大小不同，结果易得，如果大小一样，则需要进一步处理各自“编号组”，麻烦，分析如下。

考虑最简单的情况：子树与另一个子树。将它们各自的编号组表示为序列A[1...X]. 找到第一个不匹配的ID，较小那个为优。最先考虑它们各自的minID。拥有相同的minID的子树的根是minID的祖先结点，而以minID的各个祖先作为根的子树的大小不可能一样，所以大小一样的子树minID不可能一样，因此实际上这里只需要考虑它们的minID，较小者为优。

然后考虑两种补图相互比较。这时可借助其对应的子树。两个补图与其对应的子树相加，和序列一样，则对应的子树的子序列越优，它自身的子序列就越差。

再考虑补图与子树的比较（两者不一定对应）。补图或者与子树不相交，这时仍可用minID，或者包含子树，就比较麻烦。能不能从补图与子图的差入手？根必然是差的子集，由于整棵树不是有向的，谁作为根都无所谓，不影响搜索结果，如果搜索时一直都把最小的ID作为根，就保证了所有补图的minID都比所有子树的小。

#include <iostream>#include <list>#include <vector>#include <algorithm>using namespace std;struct Node {list<int> adjacentVertexes;int minID;// 以该结点为子根的子树的最小IDint subtreeNodesCount;// 以该结点为子根的子树的结点总数};struct SubGraph {enum GraphType {WITH_ROOT, WITHOUT_ROOT};// WITHOUT_ROOT表示切除边后的子树, WITH_ROOT表示该子树的补图int inVertex, outVertex, subGraphNodesCount, minID;// inVertex, outVertex 描述切除的边// minID是切除后子树的minIDGraphType type;SubGraph() {}SubGraph(int i, int o, int s, int m, GraphType t):inVertex(i), outVertex(o), subGraphNodesCount(s),minID(m), type(t) {}};void setNodes(vector<Node> &tree, int subRoot, int father) {tree[subRoot].subtreeNodesCount=1;tree[subRoot].minID=subRoot;for (list<int>::iterator iter=tree[subRoot].adjacentVertexes.begin(); iter!=tree[subRoot].adjacentVertexes.end(); iter++)if (*iter!=father)setNodes(tree, *iter, subRoot);for (list<int>::iterator iter=tree[subRoot].adjacentVertexes.begin();iter!=tree[subRoot].adjacentVertexes.end(); iter++)if (*iter!=father) {tree[subRoot].subtreeNodesCount+=tree[*iter].subtreeNodesCount;tree[subRoot].minID=min(tree[subRoot].minID, tree[*iter].minID);}}bool better(SubGraph g, SubGraph h) {if (g.subGraphNodesCount!=h.subGraphNodesCount)return g.subGraphNodesCount>h.subGraphNodesCount;if (g.type==SubGraph::WITHOUT_ROOT) {if (h.type==SubGraph::WITHOUT_ROOT)return g.minID<h.minID;elsereturn false;}else {if (h.type==SubGraph::WITHOUT_ROOT)return true;elsereturn g.minID>h.minID;}}void searchAnswer(const vector<Node> &tree, int root, int subRoot, int father, SubGraph &g) {for(list<int>::const_iterator iter=tree[subRoot].adjacentVertexes.begin();iter!=tree[subRoot].adjacentVertexes.end(); iter++)if (*iter!=father) {if (tree[*iter].subtreeNodesCount*2<=tree[root].subtreeNodesCount) {SubGraph h1(subRoot, *iter, tree[*iter].subtreeNodesCount, tree[*iter].minID, SubGraph::WITHOUT_ROOT);if (better(h1, g))g=h1;}int tmp=tree[root].subtreeNodesCount-tree[*iter].subtreeNodesCount;if (tmp*2<=tree[root].subtreeNodesCount) {SubGraph h2(subRoot, *iter, tmp, tree[*iter].minID, SubGraph::WITH_ROOT);if (better(h2, g))g=h2;}}for(list<int>::const_iterator iter=tree[subRoot].adjacentVertexes.begin();iter!=tree[subRoot].adjacentVertexes.end(); iter++)if (*iter!=father)searchAnswer(tree, root, *iter, subRoot, g);}int main() {int n, k;cin>>n>>k;// 构造树vector<Node> tree(n+1);for (int i=1; i<=n-1; i++) {int a, b;cin>>a>>b;tree[a].adjacentVertexes.push_back(b);tree[b].adjacentVertexes.push_back(a);}int root=1;// 求解vector<int> answers;SubGraph g;// 前k-1个for (int i=1; i<=k-1; i++) {setNodes(tree, root, 0);g.subGraphNodesCount=0;searchAnswer(tree, root, root, 0, g);answers.push_back(g.subGraphNodesCount);if (g.type==SubGraph::WITH_ROOT) {tree[g.outVertex].adjacentVertexes.erase(find(tree[g.outVertex].adjacentVertexes.begin(),tree[g.outVertex].adjacentVertexes.end(), g.inVertex));// 去掉子树的补图root=tree[g.outVertex].minID;// 新的根}else {tree[g.inVertex].adjacentVertexes.erase(find(tree[g.inVertex].adjacentVertexes.begin(),tree[g.inVertex].adjacentVertexes.end(), g.outVertex));// 去掉子树}}for (vector<int>::iterator iter=answers.begin(); iter!=answers.end(); iter++)cout<<(*iter)<<" ";// 第k个if (g.type==SubGraph::WITH_ROOT)cout<<tree[g.outVertex].subtreeNodesCount<<"\n";elsecout<<tree[root].subtreeNodesCount-g.subGraphNodesCount<<"\n";return 0;}// by wbchou// Feb 18th, 2013