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:
    spark基于zookeeper的高可用异常启动分析
    cdh 系统配置文件更新 error parsing conf core-site.xml
    spark2.1消费kafka0.8的数据 Recevier && Direct
    spark2.3 消费kafka0.10数据
    hadoop3.x的HA解析
    nginx安装运维
    redhat7 升级openssh openssl
    kylin 密码错误
    Vray5 材质库 图灵炸裂版 (支持Vray3.x) + VMMP
    让3dmax第一次打开材质浏览器不要卡顿
  • 原文地址:https://www.cnblogs.com/jerryRey/p/5071528.html
Copyright © 2011-2022 走看看