zoukankan      html  css  js  c++  java
  • codeforces 609E Minimum spanning tree for each edge

    E. Minimum spanning tree for each edge
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Connected undirected weighted graph without self-loops and multiple edges is given. Graph contains n vertices and m edges.

    For each edge (u, v) find the minimal possible weight of the spanning tree that contains the edge (u, v).

    The weight of the spanning tree is the sum of weights of all edges included in spanning tree.

    Input

    First line contains two integers n and m (1 ≤ n ≤ 2·105, n - 1 ≤ m ≤ 2·105) — the number of vertices and edges in graph.

    Each of the next m lines contains three integers ui, vi, wi (1 ≤ ui, vi ≤ n, ui ≠ vi, 1 ≤ wi ≤ 109) — the endpoints of the i-th edge and its weight.

    Output

    Print m lines. i-th line should contain the minimal possible weight of the spanning tree that contains i-th edge.

    The edges are numbered from 1 to m in order of their appearing in input.

    Sample test(s)
    input
    5 7
    1 2 3
    1 3 1
    1 4 5
    2 3 2
    2 5 3
    3 4 2
    4 5 4
    output
    9
    8
    11
    8
    8
    8
    9

    保证某条边e存在的MST就是普通Kruskal把e优先到了最前面。

    先求一遍MST,如果e不再MST上,是因为形成了环,把环上除了e的最大权边去掉就好了。

    (以前的LCA:用ST来RMQ,查询O(1)

    (向祖先结点倍增其实和ST差不多,查询O(logn),维护信息灵活

    (一开始想的是树剖,复杂度稍高

    #include<bits/stdc++.h>
    using namespace std;
    
    typedef long long ll;
    
    const int N = 2e5+5, M = N*2;
    
    int pa[N], rak[N];
    int fd(int x){ return pa[x] ? pa[x] = fd(pa[x]) : x; }
    bool unite(int x,int y)
    {
        int a = fd(x), b = fd(y);
        if(a == b) return false;
        if(rak[a] < rak[b]){
            pa[a] = b;
        }
        else {
            pa[b] = a;
            if(rak[a] == rak[b]) rak[a]++;
        }
        return true;
    }
    
    int fro[N], to[N], we[N];
    
    int hd[N];
    int nx[M], ver[M], wei[M];
    int ec;
    
    void add_e(int u,int v,int w)
    {
        ver[++ec] = v;
        wei[ec] = w;
        nx[ec] = hd[u];
        hd[u] = ec;
    }
    
    int n, m;
    int *cmp_c;
    bool cmp_id(int i,int j){ return cmp_c[i] < cmp_c[j]; }
    
    int r[N];
    ll kruskal()
    {
        ll re = 0;
        int i,j;
        for(i = 1; i <= m; i++) r[i] = i;
        cmp_c = we;
        sort(r+1, r + 1 + m, cmp_id);
        //ec = 0;
        for(i = 1; i <= m; i++){
            j = r[i];
            if(unite(fro[j],to[j])){
                add_e(fro[j],to[j],we[j]);
                add_e(to[j],fro[j],we[j]);
                re += we[j];
                we[j] = 0;
            }
        }
        return re;
    }
    
    const int LOG = 19;
    
    int fa[N][LOG], mx[N][LOG];
    int dep[N];
    
    void dfs(int u,int f = 0,int fw = 0,int d = 0)
    {
        fa[u][0] = f;
        mx[u][0] = fw;
        dep[u] = d;
        for(int i = hd[u]; i; i = nx[i]) {
            int v = ver[i];
            if(v == f) continue;
            dfs(v,u,wei[i],d+1);
        }
    }
    
    int lg;
    
    int queryMx(int u,int v)
    {
        int re = 0, i;
        if(dep[u] < dep[v]) swap(u,v);
        for(i = lg; i >= 0; i--) if(dep[u] - (1<<i) >= dep[v]){
            re = max(re,mx[u][i]);
            u = fa[u][i];
        }
        if(u == v) return re;
        for(i = lg; i >= 0; i--) if(fa[u][i] != fa[v][i]){
            re = max(re,max(mx[u][i],mx[v][i]));
            u = fa[u][i];
            v = fa[v][i];
        }
        return max(re,max(mx[u][0],mx[v][0]));
    }
    
    //#define LOCAL
    int main()
    {
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif
        //cout<<log2(N);
        scanf("%d%d",&n,&m);
        int i,j;
        for(i = 1; i <= m; i++){
            scanf("%d%d%d",fro+i,to+i,we+i);
        }
        ll mst = kruskal();
    
        dfs(1);
        lg = ceil(log2(n));
        for(j = 1; j <= lg; j++){
            for(i = 1; i <= n; i++) if(fa[i][j-1]){
                fa[i][j] = fa[fa[i][j-1]][j-1];
                mx[i][j] = max(mx[i][j-1],mx[fa[i][j-1]][j-1]);
            }
        }
        for(i = 1; i <= m; i++) {
            printf("%I64d
    ",we[i]?mst + we[i] - queryMx(fro[i],to[i]):mst);
        }
        return 0;
    }
  • 相关阅读:
    概念
    Jquery和Aspnet前台控件及后台代码交互
    未能找到引用的组件“Microsoft.Office.Core
    C#操作Excel,调用ApplicationClass.Quit()关闭Excel时,发生异常:Microsoft Office Word 遇到问题需要关闭
    Javasrcipt捕获按键
    使用Interop.Excel生成Excel
    Javasrcipt时间相关函数
    (转)各种纹理贴图技术
    (转)立体纹理
    (转)地形碰撞高度计算
  • 原文地址:https://www.cnblogs.com/jerryRey/p/5071528.html
Copyright © 2011-2022 走看看