SRM 621 D1L2: TreesAnalysis，DFS

题目：http://community.topcoder.com/stat?c=problem_statement&pm=13086&rd=15854

看到tree应当想到dfs，用两个dfs来标记节点。

代码如下：

#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#define FOREACH(it, X) for(__typeof((X).begin()) it = (X).begin(); it != (X).end(); ++it)/*************** Program Begin **********************/class TreesAnalysis {public:vector <int> T1[4001], T2[4001];int cntA[4001], cntB[4001];bool visited[4001], isA[4001];void dfs(int v, int v1, int v2){visited[v] = true;isA[v] = true;for (int i = 0; i < T1[v].size(); i++) {int a = T1[v][i];if (visited[a]) {continue;}if ((v == v1 && a == v2) || (v == v2 && a == v1) ) {continue;}dfs(a, v1, v2);}}void dfs2(int v){visited[v] = true;if (isA[v]) {++cntA[v];} else {++cntB[v];}for (int i = 0; i < T2[v].size(); i++) {int a = T2[v][i];if (visited[a]) {continue;}dfs2(a);cntA[v] += cntA[a];cntB[v] += cntB[a];}}long long treeSimilarity(vector <int> tree1, vector <int> tree2) {long long res = 0;int n = tree1.size() + 1;for (int i = 0; i < n - 1; i++) {T1[i].push_back(tree1[i]);T1[tree1[i]].push_back(i);T2[i].push_back(tree2[i]);T2[tree2[i]].push_back(i);}for (int i = 0; i < n - 1; i++) {int v1 = i, v2 = tree1[i];memset(visited, 0, sizeof(visited));memset(isA, 0, sizeof(isA));dfs(v1, v1, v2);int sumA = accumulate(isA, isA + n, 0);memset(visited, 0, sizeof(visited));memset(cntA, 0, sizeof(cntA));memset(cntB, 0, sizeof(cntB));dfs2(0);for (int j = 0; j < n - 1; j++) {int v3 = j, v4 = tree2[j];long long x = 0;long long a, b, c, d;if (cntA[v3] + cntB[v3] > cntA[v4] + cntB[v4]) {a = cntA[v4];b = cntB[v4];c = sumA - cntA[v4];d = n - a - b - c;} else {a = cntA[v3];b = cntB[v3];c = sumA - cntA[v3];d = n - a - b - c;}x = max(a, b);x = max(x, c);x = max(x, d);res += x * x;}}return res;}};/************** Program End ************************/