UVa Problem 10160 Servicing Stations （服务站）

// Servicing Stations （服务站）// PC/UVa IDs: 110804/10160, Popularity: B, Success rate: low Level: 3// Verdict: Accepted // Submission Date: 2011-08-10// UVa Run Time: 0.056s// 版权所有（C）2011，邱秋。metaphysis # yeah dot net//// 该题目的实质是求图的最小支配集顶点个数。求图的最小支配集问题是 NP 完全问题，目前无高效的算法。// 既然要枚举集合的所有子集，那就枚举吧，但是等一等，枚举需要个顺序，先枚举元素数量小的子集，要不// 然，先枚举元素个数大的子集可能会浪费时间，毕竟从元素个数小的子集枚举到元素个数大的子集可以保证// 找到最小支配集而不再浪费更多的时间，哎呀，时间是金钱，不是吗？可能什么时候某个强人发现 NP 难问// 题的 P 时间算法，或者证明 P = NP 那就好了，希望在有生之年能够看到！！//// 为了提高效率，压缩程序运行时间，采用了以下优化方法：// （1）若图可以拆分为多个不相连的子图，则先予拆分，然后对子图求最小支配集的顶点个数相加(之前由//     于未拆分图，导致了多次 TLE）。// （2）对于求两个集合的并采用了位操作，事先将某个顶点的邻接表表示为一个整数以便用与操作来代替集合//     的并。// （3）枚举时，先考虑度数大的顶点。//// 枚举方法参考了 [J. Loughry, J.I. van Hemert, L. Schoofs, Efficiently Enumerating// the Subsets of a Set, 2000]#include <iostream>#include <vector>#include <algorithm>#include <set>using namespace std;#define MAX_TOWN 35typedef long long unsigned LLUINT;bool finish;int minimum;vector < LLUINT > edge;LLUINT target_tag, origin_tag;// 检查顶点集合是否为支配集。void check(int flag[], int position){int index = 0;LLUINT new_tag = origin_tag, old_tag = origin_tag;for (int i = 0; i < edge.size(); i++){if ((index < position) && (flag[index] == i)){new_tag |= edge[i];if (new_tag > old_tag)old_tag = new_tag;elsereturn;index++;}}if (new_tag == target_tag){minimum = index;finish = true;}}// 按 Banker’s Sequence 枚举图的子集。void generate(int flag[], int position, int positions){if (finish)return;if (position < positions){if (position == 0){for (int i = 0; i < edge.size(); i++){flag[position] = i;generate(flag, position + 1, positions);}}else{for (int i = flag[position - 1] + 1; i < edge.size(); i++){flag[position] = i;generate(flag, position + 1, positions);}}}elsecheck(flag, position);}// 枚举图向量的子集以判断是否是一个支配集。void enumerating_subset(){for (int i = 1; i <= edge.size(); i++){int * flag = new int[edge.size()];generate(flag, 0, i);delete [] flag;if(finish)return;}}bool cmp(LLUINT x, LLUINT y){return x > y;}// 获取图的最小支配集顶点数（MDSN）。int mdsn(vector < vector < int > > &vertex){int base = 0;origin_tag = 0;target_tag = 0;vector < bool > dirty(vertex.size());fill(dirty.begin(), dirty.end(), false);// 清掉度为 0 的点。度为 0 的点和其他点都无通路，则该图的最小支配集必须要包括// 该顶点。则表示最小支配集顶点个数的变量 base 需增加 1。for (int i = 0; i < vertex.size(); i++){if (vertex[i].size() == 0){base++;origin_tag |= ((LLUINT)1 << i);}if (vertex[i].size() == 1 && dirty[i] == false){dirty[i] = true;if (dirty[vertex[i][0] - 1] == false){base++;dirty[vertex[i][0] - 1] = true;}}target_tag |= ((LLUINT)1 << i);}// 清掉度为 1 的点。度为 1 的点表明该顶点 A 只与其他一个顶点 B 相连接，则可// 将 B 计入最小支配集中。edge.clear();for (int i = 0; i < vertex.size(); i++){if (dirty[i] == true){origin_tag |= ((LLUINT)1 << i);for (int j = 0; j < vertex[i].size(); j++)origin_tag |= ((LLUINT)1 << (vertex[i][j] - 1));}if (dirty[i] == false && vertex[i].size() > 0){LLUINT t = ((LLUINT)1 << i);for (int j = 0; j < vertex[i].size(); j++)t |= ((LLUINT)1 << (vertex[i][j] - 1));edge.push_back(t);}}// 排序，度数大的点首先考虑。sort(edge.begin(), edge.end(), cmp);minimum = 0;finish = false;enumerating_subset();return (base + minimum);}int servicing_stations(vector < vector < int > > &vertex){// 使用宽度优先搜索分离子图，计算子图的最小支配集顶点个数，相// 加即为原图的最小支配集顶点个数。int total = 0;while (vertex.size() > 0){vector < vector < int > > open;set < int > close;int size = 0;open.push_back(vertex[0]);close.insert(vertex[0][0]);vertex.erase(vertex.begin());while (open.size() > size){int origin = size;int current = open.size() - 1;size = open.size();for (int i = origin; i <= current; i++){for (int j = 1; j < open[i].size(); j++){if (close.find(open[i][j]) == close.end()){close.insert(open[i][j]);for (int m = 0; m < vertex.size(); m++){if (vertex[m][0] == open[i][j]){open.push_back(vertex[m]);vertex.erase(vertex.begin() + m);break;}}}}}}// 调整分离出的子图的序号以便后续操作。vector < vector < int > > tmp;for (int c = 1; c <= MAX_TOWN; c++)if (close.find(c) != close.end()){for (int i = 0; i < open.size(); i++)if (open[i][0] == c)tmp.push_back(open[i]);}for (int i = 0; i < tmp.size(); i++){int current = tmp[i][0];for (int m = 0; m < tmp.size(); m++)for (int n = 1; n < tmp[m].size(); n++)if (tmp[m][n] == current)tmp[m][n] = (i + 1);}for (int i = 0; i < tmp.size(); i++)tmp[i].erase(tmp[i].begin());// 计算该子图的最小支配集顶点数。total += mdsn(tmp);}return total;}int main(int ac, char *av[]){int n;int m;int x, y;vector < vector < int > > vertex;while (cin >> n >> m, n && m){vertex.clear();vertex.resize(n);// 数组的第一个数存放顶点的序号。for (int i = 0; i < n; i++)vertex[i].push_back((i + 1));// 读入镇之间的通路。for (int i = 0; i < m; i++){// 自身连接到自身的通路不添加。cin >> x >> y;if (x != y){vertex[x - 1].push_back(y);vertex[y - 1].push_back(x);}}cout << servicing_stations(vertex) << endl;}return 0;}