TCO14 2C L2: CliqueGraph，graph theory, clique

称号：http://community.topcoder.com/stat?c=problem_statement&pm=13251&rd=16017

參考：http://apps.topcoder.com/wiki/display/tc/TCO+2014+Round+2C

假设用先计算出每条边，用邻接矩阵来表示图，然后用BFS或 Floyd-Warshall算法来计算距离的话。时间复杂度是O(N^3)，会超时。依据题名的提示知要利用clique graph的性质来做。基本思想是在BFS的时候将一个clique看成一个总体。一旦訪问到clique中的一个点，则这个clique中全部点的距离都能够得到。算法描写叙述例如以下

for each source vertex s:
    mark all vertices and all cliques as unvisited
    start BFS from s
    when processing a vertex v during the BFS:
        for each unvisited clique C that contains v:
            mark C as visited
            use edges in C to discover new vertices

代码：

#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <bitset>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>
using namespace std;

#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)
typedef pair<int, int> pii;
typedef long long llong;
typedef pair<llong, llong> pll;
#define mkp make_pair

/*************** Program Begin **********************/
bool visited_vertex[5001];
bool visited_clique[5001];


class CliqueGraph {
public:
	long long calcSum(int N, vector <int> V, vector <int> sizes) {
		long long res = 0;
		vector <int> S(sizes.size() + 1);
		S[0] = 0;
		for (int i = 0; i < sizes.size(); i++) {
			S[i + 1] += S[i] + sizes[i];
		}
		vector <vector<int>> cliques(sizes.size());	// clique i 包括的点
		vector <vector<int>> vcliques(N);  // 包括点v的cliques
		for (int i = 0; i < sizes.size(); i++) {
			for (int j = S[i]; j < S[i + 1]; j++) {
				cliques[i].push_back(V[j]);
				vcliques[ V[j] ].push_back(i);
			}
		}

		for (int src = 0; src < N; src++) {
			vector <int> D(N, 123456789);
			D[src] = 0;
			memset(visited_vertex, 0, sizeof(visited_vertex));
			memset(visited_clique, 0, sizeof(visited_clique));
			queue <int> Q;
			Q.push(src);
			visited_vertex[src] = true;
			while (!Q.empty()) {
				int v = Q.front();
				Q.pop();
				// 更新全部包括v的cliques中的全部点
				for (int i = 0; i < vcliques[v].size(); i++) {
					int c = vcliques[v][i];	// 包括v的cliques
					if (visited_clique[c]) {
						continue;
					}
					visited_clique[c] = true;
					for (int j = 0; j < cliques[c].size(); j++) {
						int u = cliques[c][j];	// clique c z中的的点
						if (visited_vertex[u]) {
							continue;
						}
						visited_vertex[u] = true;
						D[u] = D[v] + 1;
						Q.push(u);
					}
				}
			}
			for (int i = 0; i < N; i++) {
				res += D[i];
			}
		}
		return res / 2;
	};
};

/************** Program End ************************/