zoukankan      html  css  js  c++  java
  • 【CF617D】Roads in Yusland

    【CF617D】Roads in Yusland

    题面

    description

    蒯的洛谷的

    题解

    我们现在已经转化好了题目了,戳这里

    那么我们考虑怎么求这个东西,我们先判断一下是否所有的边都能被覆盖,不行的话输出(-1)

    再将路径(u ightarrow v(dep_u>dep_v))以其权值为关键字丢到以(u)为根的左偏树(小根)上,

    再进行(dfs),合并(x)的所有儿子的左偏树,对于点(x),我们能选则选,

    还有打标记等细节,详见代码。

    代码

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring> 
    #include <cmath> 
    #include <algorithm>
    using namespace std; 
    namespace IO { 
        const int BUFSIZE = 1 << 20; 
        char ibuf[BUFSIZE], *is = ibuf, *it = ibuf; 
        inline char gc() { 
            if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin); 
            return *is++; 
        } 
    } 
    inline int gi() {
        register int data = 0, w = 1;
        register char ch = 0;
        while (ch != '-' && (ch > '9' || ch < '0')) ch = IO::gc();
        if (ch == '-') w = -1 , ch = IO::gc();
        while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = IO::gc();
        return w * data;
    } 
    const int MAX_N = 3e5 + 5; 
    struct Graph { int to, next; } e[MAX_N << 1];
    int fir[MAX_N], e_cnt; 
    void clearGraph() { memset(fir, -1, sizeof(fir)); e_cnt = 0; } 
    void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}; fir[u] = e_cnt++; }
    struct chain { int u, v, w; } a[MAX_N]; 
    int N, M, c[MAX_N], dep[MAX_N];
    long long ans = 0; 
    void dfs(int x, int f) {
        dep[x] = dep[f] + 1; 
        for (int i = fir[x]; ~i; i = e[i].next) {
            int v = e[i].to; if (v == f) continue;
            dfs(v, x), c[x] += c[v]; 
        } 
    } 
    struct Node {
        int ls, rs, dis;
        int v, ed, tag;
    } t[MAX_N]; 
    int rt[MAX_N];
    void puttag(int x, int v) { t[x].tag += v, t[x].v += v; } 
    void pushdown(int x) {
        if (!t[x].tag) return ;
        if (t[x].ls) puttag(t[x].ls, t[x].tag); 
        if (t[x].rs) puttag(t[x].rs, t[x].tag); 
        t[x].tag = 0; 
    } 
    int merge(int x, int y) {
        if (!x || !y) return x + y; 
        pushdown(x), pushdown(y); 
        if (t[x].v > t[y].v) swap(x, y); 
        t[x].rs = merge(t[x].rs, y); 
        if (t[t[x].ls].dis < t[t[x].rs].dis) swap(t[x].ls, t[x].rs); 
        t[x].dis = t[t[x].rs].dis + 1; 
        return x; 
    } 
    void solve(int x, int f) { 
        for (int i = fir[x]; ~i; i = e[i].next)
            if (e[i].to != f) {
                solve(e[i].to, x); 
                rt[x] = merge(rt[e[i].to], rt[x]); 
            }
        if (x == 1) return ; 
        while (dep[t[rt[x]].ed] >= dep[x]) rt[x] = merge(t[rt[x]].ls, t[rt[x]].rs); 
        int w = t[rt[x]].v; 
        ans += w, puttag(rt[x], -w); 
    } 
    int main () { 
    #ifndef ONLINE_JUDGE 
        freopen("cpp.in", "r", stdin); 
    #endif
        clearGraph(); 
        N = gi(), M = gi();
        for (int i = 1; i < N; i++) {
            int u = gi(), v = gi(); 
            Add_Edge(u, v), Add_Edge(v, u); 
        } 
        for (int i = 1; i <= M; i++) a[i] = (chain){gi(), gi(), gi()}; 
        for (int i = 1; i <= M; i++) c[a[i].u]++, c[a[i].v]--; 
        dfs(1, 0); 
        for (int i = 2; i <= N; i++) if (c[i] <= 0) return puts("-1") & 0; 
        for (int i = 1; i <= M; i++) t[i] = (Node){0, 0, 0, a[i].w, a[i].v, 0}, rt[a[i].u] = merge(rt[a[i].u], i);
        solve(1, 0); 
        cout << ans << endl; 
        return 0; 
    } 
    

    跑到洛谷(rank1)纪念一下:

    img

    好像还是cf的(rank1)呢qaq:

    img1

  • 相关阅读:
    Ruby on rails3新手谈(1):Ruby on rails环境搭建
    更灵活,更易维护的WebHandler之通用webHandler编码方案(2)
    .Net Remoting之windows服务部署
    Ruby on rails3新手谈(2):简单的例子helloworld
    分组取最新记录的SQL
    验证日期的Javascript
    使用Javascript创建遮罩层并模拟Alert、Confirm对话框
    取得当前鼠标的X,Y坐标,及相关属性的介绍
    B/S结构系统中使用Session遇到的问题
    IE与FireFox的兼容性问题
  • 原文地址:https://www.cnblogs.com/heyujun/p/10520039.html
Copyright © 2011-2022 走看看