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  • cf1042F. Leaf Sets(贪心)

    题意

    题目链接

    给出一棵树,删除一些边,使得任意联通块内的任意点距离不超过$k$

    sol

    考场上想的贪心是对的:考虑一棵子树,如果该子树内最深的两个节点的距离相加$>k$就删掉最深的那个点,向上update的时候只返回最深的点的深度

    然而却苦于写不出代码。。。

    /*
    
    */
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<map>
    #include<vector>
    #include<set>
    #include<queue>
    #include<cmath>
    //#include<ext/pb_ds/assoc_container.hpp>
    //#include<ext/pb_ds/hash_policy.hpp>
    #define Pair pair<int, int>
    #define MP(x, y) make_pair(x, y)
    #define fi first
    #define se second
    #define int long long 
    #define LL long long 
    #define ull unsigned long long 
    #define rg register 
    #define pt(x) printf("%d ", x);
    //#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
    //char buf[(1 << 22)], *p1 = buf, *p2 = buf;
    //char obuf[1<<24], *O = obuf;
    //void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
    //#define OS  *O++ = ' ';
    using namespace std;
    //using namespace __gnu_pbds;
    const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
    const double eps = 1e-9;
    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, K, ans = 0;
    vector<int> v[MAXN];
    int dfs(int x, int fa) {
        if(v[x].size() == 1) return 0;
        vector<int> dis;
        for(int i = 0; i < v[x].size(); i++) {
            int to = v[x][i];
            if(to == fa) continue;
            dis.push_back(dfs(to, x) + 1);        
        }
        sort(dis.begin(), dis.end());
        while(dis.size() >= 2) {
            int x = dis[dis.size() - 1], y = dis[dis.size() - 2];
            if(x + y > K) 
                ans++, dis.pop_back();
            else break;
        }
        return dis.back();
    }
    main() {
        N = read(); K = read();
        for(int i = 1; i <= N - 1; i++) {
            int x = read(), y = read();
            v[x].push_back(y);
            v[y].push_back(x);
        }
        for(int i = 1; i <= N; i++)
            if(v[i].size() > 1) {dfs(i, 0); break;}
        printf("%d", ans + 1);
        return 0;
    }
    /*
    2 2 1
    1 1
    2 1 1
    */
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/9664498.html
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