【并查集缩点+tarjan无向图求桥】Where are you @牛客练习赛32 D

目录

• 【并查集缩点+tarjan无向图求桥】Where are you @牛客练习赛32 D
  • PROBLEM
  • SOLUTION
  • CODE

【并查集缩点+tarjan无向图求桥】Where are you @牛客练习赛32 D

PROBLEM

时空裂隙

SOLUTION

楠神口胡算法我来实现系列

从小到大枚举边权，对于当前的权值，在当前的图找出所有等于该权值的边，把这些边的顶点用其在并查集中的代表元（即fa[x]）替换，然后建图，求所建图的桥边。求完之后把每条边的两个顶点合并（缩点），然后枚举下一个权值。最后统计桥边数量和就是答案。

CODE

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 2e5 + 5;

struct edge {
    int v, w, nex;
} ed[MAXN * 2];

int head[MAXN], tot;

void addedge(int u, int v,int w) {
    tot++;
    ed[tot].v = v;
    ed[tot].w = w;
    ed[tot].nex = head[u];
    head[u] = tot;
}

set<int> ss;

struct data{
    int ind,w;
}eg[MAXN*2];

bool cmp(data a,data b){
    return a.w<b.w;
}

int n,m;

bool bridge[MAXN*2];
int dfn[MAXN],low[MAXN],num;
vector<pair<int,int>> g[MAXN];
set<int> nd;

int fa[MAXN];

int DjsGet(int x){
    if(x==fa[x])return x;
    return fa[x] = DjsGet(fa[x]);
}

void tarjan(int x,int in_edge){
    dfn[x] = low[x] = ++num;
    for(auto ver:g[x]){
        int y = ver.first;
        int i = ver.second;
        if(!dfn[y]){
            tarjan(y,i);
            low[x] = min(low[x],low[y]);
            if(low[y]>dfn[x]){
                bridge[i] = bridge[i^1] = true;
            }
        }
        else if(i!=(in_edge^1))
            low[x] = min(low[x],dfn[y]);
    }
}

int main() {
    scanf("%d%d%*d",&n,&m);
    tot = 1;
    for(int i=1;i<=m;i++){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        addedge(u,v,w);
        addedge(v,u,w);
        ss.insert(w);
        eg[i]= {i*2,w};
    }
    sort(eg+1,eg+m+1,cmp);//边按权值从小到大排序
    for(int i = 1;i<=n;i++)fa[i] = i;
    int pos = 1;
    for(auto curval:ss){
        int pre = pos;
        //建子图
        for(pos;eg[pos].w<=curval&&pos<=m;pos++){
            int ind = eg[pos].ind;
            int x = ed[ind^1].v,y = ed[ind].v;
            int fx = DjsGet(x);
            int fy = DjsGet(y);
            if(fx==fy)continue;
            g[fx].push_back(make_pair(fy,ind));
            g[fy].push_back(make_pair(fx,ind^1));
            nd.insert(fx);
            nd.insert(fy);
        }
        //求桥
        for(auto i:nd){
            if(!dfn[i])tarjan(i,0);
        }
        //init
        for(auto i:nd){
            dfn[i] = low[i] = 0;
            g[i].clear();
        }
        num = 0;
        nd.clear();
        //缩点
        for(int i= pre;i<pos;i++){
            int ind = eg[i].ind;
            int x = ed[ind^1].v,y = ed[ind].v;
            int fx = DjsGet(x);
            int fy = DjsGet(y);
            if(fx==fy)continue;
            fa[fx] = fy;
        }
    }
    int ans = 0;
    for(int i = 2;i<=tot;i+=2){
        if(bridge[i])ans++;
    }
    printf("%d\n",ans);
    return 0;
}