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\n",we[i]?mst + we[i] - queryMx(fro[i],to[i]):mst);
    }
    return 0;
}