UVALive 6837 (最小生成树)

最新推荐文章于 2019-09-10 17:58:25 发布

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本文详细阐述了如何通过求解任意一棵最小生成树，并标记树边和非树边，进而通过枚举非树边来确定哪些边是不可替代的，以及计算这些边的总权重。

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题意是给定一个联通图,求这个图的最小生成树的不可替代边有哪些,并计算这些边的总权值.

先求出任意一棵MST,然后标记树边和非树边,然后枚举非树边,对于非树边的两个点u,v,在MST中如果u->v的路径上有边的权值等于这条非树边的权值那么这条树边就是可替代边,给它打上标记.u->v的路径可以通过u->LCA(u,v), v->LCA (u,v)暴力枚举.

复杂度(m*n).

<span style="font-size:12px;">#include <bits/stdc++.h>
using namespace std;
#define maxn 1111
#define maxm 511111
#define find Find

struct node1 {
    int u, v, w;
    bool vis;
    bool operator < (node1 a) const {
        return w < a.w;
    }
}e[maxm];
struct node2 {
    int u, v, w, next;
    bool vis;
}edge[maxm];

int head[maxn], fa[maxn], n, m;
int cnt;
int MST;

int find (int x) {
    return fa[x] == x ? fa[x] : fa[x] = find (fa[x]);
}

void add_edge (int u, int v, int w) {
    edge[cnt].vis = 0;
    edge[cnt].u = u, edge[cnt].v = v, edge[cnt].w = w, edge[cnt].next = head[u], head[u] = cnt++;
}

void kruscal () {
    for (int i = 1; i <= n; i++)
        fa[i] = i;
    int num = 0;
    MST = 0;
    for (int i = 1; i <= m; i++) {
        int u = e[i].u, v = e[i].v, w = e[i].w;
        int p1 = find (u), p2 = find (v);
        if (p1 != p2) {
            fa[p1] = p2;
            add_edge (u, v, w);
            add_edge (v, u, w);
            num++;
            MST += e[i].w;
            e[i].vis = 1;
        }
        if (num == n-1)
            break;
    }
}

int d[maxn];
void dfs (int u, int fa, int deep) {
    d[u] = deep;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].v;
        if (v == fa)
            continue;
        dfs (v, u, deep+1);
    }
}

void work (int u, int v, int w) {
    if (u == v) {
        return ;
    }
    if (d[u] < d[v])
        swap (u, v);
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int to = edge[i].v;
        if (d[to] > d[u])
            continue ;
        if (edge[i].w == w) {
            edge[i].vis = edge[i^1].vis = 1;
        }
        work (to, v, w);
    }
}

void solve () {
    kruscal ();
    dfs (1, 0, 1);
    for (int i = 1; i <= m; i++) {
        if (!e[i].vis) {
            work (e[i].u, e[i].v, e[i].w);
        }
    }
    int ans = 0;
    for (int i = 0; i < cnt; i += 2) {
        if (edge[i].vis)
            MST -= edge[i].w;
        else ans++;
    }
    printf ("%d %d\n", ans, MST);
}

int main () {
    //freopen ("in", "r", stdin);
    while (scanf ("%d%d", &n, &m) == 2) {
        for (int i = 1; i <= m; i++) {
            int u, v, w;
            scanf ("%d%d%d", &u, &v, &w);
            e[i].u = u, e[i].v = v, e[i].w = w;
            e[i].vis = 0;
        }
        memset (head, -1, sizeof head);
        cnt = 0;
        sort (e+1, e+1+m);
        solve ();
    }
    return 0;
}
</span>