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  • bzoj 1116

    思路:每个连通块都判是否有环。

    #include<bits/stdc++.h>
    #define LL long long
    #define fi first
    #define se second
    #define mk make_pair
    #define pii pair<int,int>
    #define piii pair<int, pair<int,int> >
    
    using namespace std;
    
    const int N = 1e5 + 10;
    const int M = 10 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-6;
    
    int n, m, vis[N];
    vector<int> edge[N];
    
    bool dfs(int u, int p) {
        vis[u] = -1;
        for(int i = 0; i < edge[u].size(); i++) {
            int v = edge[u][i];
            if(v == p) continue;
            if(vis[v] == -1) {
                return false;
            } else if(!vis[v] && !dfs(v, u)) {
                return false;
            }
        }
        vis[u] = 1;
        return true;
    }
    int main() {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= m; i++) {
            int u, v; scanf("%d%d", &u, &v);
            edge[u].push_back(v);
            edge[v].push_back(u);
        }
    
        for(int i = 1; i <= n; i++) {
            if(!vis[i] && dfs(i, 0)) {
                puts("NIE");
                return 0;
            }
        }
    
        puts("TAK");
        return 0;
    }
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/9187079.html
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