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  • Codeforces VK Cup 2012 Round 3 A. Variable, or There and Back Again(dfs)

    题目链接:http://codeforces.com/problemset/problem/164/A

    思路:用vector分别保留原图和发图,然后分别从val值为1的点正向遍历,va值为2的点反向遍历,如果某个点这两种方式都可以遍历到,则输出1,否则输出0.

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    #define REP(i, a, b) for (int i = (a); i < (b); ++i)
    #define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
    using namespace std;
    
    const int MAX_N = (100000 + 100);
    int N, M, val[MAX_N], vis1[MAX_N], vis2[MAX_N];
    int path[MAX_N];
    vector<int > g1[MAX_N], g2[MAX_N];
    
    void dfs(int u, int fa)
    {
    	vis1[u] = 1;
    	REP(i, 0, (int)g1[u].size()) {
    		int v = g1[u][i];
    		if (!vis1[v] && v != fa && val[v] != 1) dfs(v, u);
    	}
    }
    
    void rdfs(int u, int fa)
    {
    	vis2[u] = 1;
    	if (val[u] == 1) return;
    	REP(i, 0, (int)g2[u].size()) {
    		int v = g2[u][i];
    		if (!vis2[v] && v != fa) rdfs(v, u);
    	}
    }
    
    
    
    int main()
    {
    	cin >> N >> M;
    	FOR(i, 1, N) cin >> val[i];
    	FOR(i, 1, M) {
    		int u, v; cin >> u >> v;
    		g1[u].push_back(v);
    		g2[v].push_back(u);
    	}
    	FOR(i, 1, N)  {
    		if (val[i] == 1) dfs(i, -1);
    		else if (val[i] == 2) rdfs(i, -1);
    	}
    	FOR(i, 1, N) printf("%d
    ", (vis1[i] & vis2[i]));
    	return 0;
    }

    
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  • 原文地址:https://www.cnblogs.com/wally/p/4477063.html
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