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  • cf547D. Mike and Fish(欧拉回路)

    题意

    题目链接

    Sol

    说实话这题我到现在都不知道咋A的。

    考试的时候是对任意相邻点之间连边,然后一分没有

    然后改成每两个之间连一条边就A了。。

    按说是可以过掉任意坐标上的点都是偶数的数据啊。。

    #include<cstdio>
    #include<algorithm>
    #include<iostream>
    #include<vector>
    #include<cstring>
    #include<queue>
    #define Pair pair<int, int>
    #define MP(x, y) make_pair(x, y)
    #define fi first
    #define se second 
    using namespace std;
    const int MAXN = 2e5 + 10, INF = 1e9 + 10, mod = 998244353;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N;
    int x[MAXN], y[MAXN], date[MAXN], numx[MAXN], numy[MAXN];
    vector<int> E[MAXN];
    vector<Pair> tmp[MAXN];
    void Get(int *a) {
        for(int i = 1; i <= N; i++) date[i] = a[i];
        sort(date + 1, date + N + 1);
        int num = unique(date + 1, date + N + 1) - date - 1;
        for(int i = 0; i <= num; i++) tmp[i].clear();
        for(int i = 1; i <= N; i++) a[i] = lower_bound(date + 1, date + num + 1, a[i]) - date - 1, tmp[a[i]].push_back(MP(a[i], i));
        for(int i = 0; i <= num; i++) {
            if(tmp[i].size() < 2) continue;
            sort(tmp[i].begin(), tmp[i].end());
            for(int j = 0; j < tmp[i].size() - 1; j += 2) {
                int x = tmp[i][j].se, y = tmp[i][j + 1].se;
                E[x].push_back(y);
                E[y].push_back(x);
            }
        }
    }
    int vis[MAXN], vis2[MAXN];
    void BFS(int k) {
        queue<int> q;
        vis[k] = 1; q.push(k); 
        while(!q.empty()) {
            int p = q.front(); q.pop(); 
            if(vis2[p]) continue; vis2[p] = 1;
            for(int i = 0; i < E[p].size(); i++) {
                int to = E[p][i];
                vis[to] = vis[p] ^ 1;
                q.push(to);
            }
        }
        
    }
    int main() {
        N = read();
        for(int i = 1; i <= N; i++) x[i] = read(), y[i] = read();
        Get(x); Get(y);
        for(int i = 1; i <= N; i++) if(!vis2[i]) BFS(i);
        for(int i = 1; i <= N; i++) putchar(vis[i] == 0 ? 'b' : 'r');
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/9839468.html
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