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