codeforces 1108F MST Unification 次小生成树 lca

F. MST Unification

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an undirected weighted connected graph with nn vertices and mm edges without loops and multiple edges.

The ii-th edge is ei=(ui,vi,wi)ei=(ui,vi,wi); the distance between vertices uiui and vivi along the edge eiei is wiwi (1≤wi1≤wi). The graph is connected, i. e. for any pair of vertices, there is at least one path between them consisting only of edges of the given graph.

A minimum spanning tree (MST) in case of positive weights is a subset of the edges of a connected weighted undirected graph that connects all the vertices together and has minimum total cost among all such subsets (total cost is the sum of costs of chosen edges).

You can modify the given graph. The only operation you can perform is the following: increase the weight of some edge by 11. You canincrease the weight of each edge multiple (possibly, zero) times.

Suppose that the initial MST cost is kk. Your problem is to increase weights of some edges with minimum possible number of operationsin such a way that the cost of MST in the obtained graph remains kk, but MST is unique (it means that there is only one way to choose MST in the obtained graph).

Your problem is to calculate the minimum number of operations required to do it.

Input

The first line of the input contains two integers nn and mm (1≤n≤2⋅105,n−1≤m≤2⋅1051≤n≤2⋅105,n−1≤m≤2⋅105) — the number of vertices and the number of edges in the initial graph.

The next mm lines contain three integers each. The ii-th line contains the description of the ii-th edge eiei. It is denoted by three integers ui,viui,vi and wiwi (1≤ui,vi≤n,ui≠vi,1≤w≤1091≤ui,vi≤n,ui≠vi,1≤w≤109), where uiui and vivi are vertices connected by the ii-th edge and wiwi is the weight of this edge.

It is guaranteed that the given graph doesn't contain loops and multiple edges (i.e. for each ii from 11 to mm ui≠viui≠vi and for each unordered pair of vertices (u,v)(u,v) there is at most one edge connecting this pair of vertices). It is also guaranteed that the given graph is connected.

Output

Print one integer — the minimum number of operations to unify MST of the initial graph without changing the cost of MST.

Examples

input

Copy

8 10
1 2 1
2 3 2
2 4 5
1 4 2
6 3 3
6 1 3
3 5 2
3 7 1
4 8 1
6 2 4

output

Copy

1

input

Copy

4 3
2 1 3
4 3 4
2 4 1

output

Copy

0

input

Copy

3 3
1 2 1
2 3 2
1 3 3

output

Copy

0

input

Copy

3 3
1 2 1
2 3 3
1 3 3

output

Copy

1

input

Copy

1 0

output

Copy

0

input

Copy

5 6
1 2 2
2 3 1
4 5 3
2 4 2
1 4 2
1 5 3

output

Copy

2

给定一个图，一个操作为给一条边权值加一，求最少多少次操作使得最小生成树是唯一的。

即使得次小生成树的权重大于最小生成树，因此生成任意一个最小生成树之后，枚举添加任意不在该最小生成树上的边，此时会形成环，如果该环上最大值的边等于新添加进来的边，那么最小生成树不是唯一的，就需要在新加上的边上进行一次操作。

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 2e5 + 10;
struct edge {
    int u, v, w, use;
};
struct eop {
    bool operator() (const edge& a, const edge& b)const {
        return a.w < b.w;
    }
};
vector<edge> es;
vector<edge> G[N];
int n, m, p[N], fa[N][25], mcost[N][25], cost[N], depth[N];

int find(int v) {
    if (p[v] == v) return v;
    return p[v] = find(p[v]);
}

void dfs(int u, int f) {
    depth[u] = depth[f] + 1;
    p[u] = f;
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i].v;
        int w = G[u][i].w;
        if (v == f) continue;
        cost[v] = w;
        //printf("---%d %d\n", v, u);
        dfs(v, u);
    }
}

void process() {
    for (int i = 1; i <= n; i++) {
        fa[i][0] = p[i];
        mcost[i][0] = cost[i];
    }
    for (int k = 1; (1 << k) <= n; k++)
        for (int i = 1; i <= n; i++) {
            if (fa[i][k - 1] != 0) {
                fa[i][k] = fa[fa[i][k - 1]][k - 1];
                mcost[i][k] = max(mcost[i][k - 1], mcost[fa[i][k - 1]][k - 1]);
            }
        }
}

int query(int u, int v) {
    int ans = 0;
    if (depth[u] < depth[v]) swap(u, v);
    for (int i = 20; i >= 0; i--) {
        if (depth[u] - (1 << i) >= depth[v]) {
            ans = max(ans, mcost[u][i]);
            u = fa[u][i];
        }
    }
    if (u == v) return ans;
    for (int i = 20; i >= 0; i--) {
        if (fa[u][i] != 0 && fa[u][i] != fa[v][i]) {
            ans = max(mcost[u][i], ans);
            ans = max(mcost[v][i], ans);
            u = fa[u][i]; v = fa[v][i];
        }
    }
    ans = max(ans, cost[u]);
    ans = max(ans, cost[v]);
    return ans;
}

int main() {
    scanf("%d%d", &n, &m);
    int u, v, w;
    memset(fa, 0, sizeof(fa));
    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d", &u, &v, &w);
        es.push_back({u, v, w, 0});
    }
    for (int i = 1; i <= n; i++) p[i] = i;
    sort(es.begin(), es.end(), eop());
    for (int i = 0; i < es.size(); i++) {
        u = es[i].u; v = es[i].v, w = es[i].w;
        int fu = find(u), fv = find(v);
        if (fu != fv) {
            p[fu] = fv;
            G[u].push_back({u, v, w, 0});
            G[v].push_back({v, u, w, 0});
            es[i].use = 1;
        }
    }
    depth[0] = 0;
    cost[1] = 0;
    dfs(1, 0);
    process();

    int ans = 0;
    for (int i = 0; i < es.size(); i++) {
        if (es[i].use) continue;
       // printf ("%d %d %d %d\n", es[i].u, es[i].v, es[i].w, query(es[i].u, es[i].v));
        if (query(es[i].u, es[i].v) == es[i].w) ans ++;
    }
    printf("%d\n", ans);
    return 0;
}