概述
For a connected undirected weighted graph G, MST (minimum spanning tree) is a subgraph of G that contains all of G's vertices, is a tree, and sum of its edges is minimum possible.
You are given a graph G. If you run a MST algorithm on graph it would give you only one MST and it causes other edges to become jealous. You are given some queries, each query contains a set of edges of graph G, and you should determine whether there is a MST containing all these edges or not.
The first line contains two integers n, m (2 ≤ n, m ≤ 5·105, n - 1 ≤ m) — the number of vertices and edges in the graph and the number of queries.
The i-th of the next m lines contains three integers ui, vi, wi (ui ≠ vi, 1 ≤ wi ≤ 5·105) — the endpoints and weight of the i-th edge. There can be more than one edges between two vertices. It's guaranteed that the given graph is connected.
The next line contains a single integer q (1 ≤ q ≤ 5·105) — the number of queries.
q lines follow, the i-th of them contains the i-th query. It starts with an integer ki (1 ≤ ki ≤ n - 1) — the size of edges subset and continues with ki distinct space-separated integers from 1 to m — the indices of the edges. It is guaranteed that the sum of ki for 1 ≤ i ≤ q does not exceed 5·105.
For each query you should print "YES" (without quotes) if there's a MST containing these edges and "NO" (of course without quotes again) otherwise.
5 7 1 2 2 1 3 2 2 3 1 2 4 1 3 4 1 3 5 2 4 5 2 4 2 3 4 3 3 4 5 2 1 7 2 1 2
YES NO YES NO
This is the graph of sample:
Weight of minimum spanning tree on this graph is 6.
MST with edges (1, 3, 4, 6), contains all of edges from the first query, so answer on the first query is "YES".
Edges from the second query form a cycle of length 3, so there is no spanning tree including these three edges. Thus, answer is "NO".
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const int N=501000;
int n,m,u[N],v[N],w[N],f[N],p[N],qrcase[N],T,wa[N];
int Q,k,id,x,maxw;
vector<int> eg[N];
vector<PII> qw[N];
int nxt1[N], nxt2[N];
int find1(int now){
if (nxt1[now]==now) return now;
return nxt1[now]=find1(nxt1[now]);
}
int find2(int now){
if (qrcase[now]!=T){qrcase[now]=T; nxt2[now]=nxt1[now];}
if (nxt2[now]==now) return now;
return nxt2[now]=find2(nxt2[now]);
}
int main() {
maxw=0;
scanf("%d%d",&n,&m);
for (int i=1; i<=m; i++){
scanf("%d%d%d", u+i, v+i, w+i);
eg[w[i]].push_back(i);
maxw=max(maxw, w[i]);
}
scanf("%d", &Q);
for (int i=0; i<Q; i++){
scanf("%d", &id);
for (int j=0; j<id; j++){
scanf("%d", &x);
qw[w[x]].push_back(mp(i, x));
}
}
memset(wa, 0, sizeof(wa));
for (int i=1; i<=n; i++){
nxt1[i]=i;
}
T=0;
for (int i=1; i<=maxw; i++){
sort(qw[i].begin(), qw[i].end());
for (int j=0; j<qw[i].size(); j++){
if (j==0 || qw[i][j].first!=qw[i][j-1].first) T++;
int x=u[qw[i][j].second], y=v[qw[i][j].second];
if (find2(x)==find2(y)){
wa[qw[i][j].first]=true;
}
nxt2[find2(y)]=nxt2[find2(x)];///nxt2[y]=x;
}
for (int j=0; j<eg[i].size(); j++){
int x=find1(u[eg[i][j]]); int y=find1(v[eg[i][j]]);
nxt1[find1(y)]=find1(x);
}
}
for (int i=0; i<Q; i++){
if (wa[i]) printf("NOn"); else printf("YESn");
}
}
两个注意点
第一点是原理，按照边权值由小到大的顺序，处理每个query的边。用两个并查集，第一个并查集next1是原图所有边，第二个是每个query的边。保证处理query边的时候原图的权值小的边已经处理过了，如图1，这样一旦有回路说明可以用小的边来替换当前query的边，就不存在了，不然就是存在的！
第二点，注意并查集的next数组的更新。
再传一个别人写的
题目大意： 给出一个n个顶点， m条边的联通无向图， 每条边有权值 wi ，有q组询问， 每次询问原图中的 ki 条边， 求是否存在一种最小生成树包含这 ki 条边。 (n,m,wi,q,∑ki≤5∗105) 。
思路： 考虑离线求解。 先将所有边按从小到大排序， 每次同时考虑一堆权值均为x的边， 根据kruskal算法， 所有 < x的边已经进行了并查集缩点， 再单独考虑所有涉及了这一次权值为x的询问
qi
， 把
qi
涉及到的边单独拿出来做并查集合并， 若出现了环则
qi
这条询问判断为不存在， 再把这次的修改还原， 考虑完所有的涉及到的询问后， 把该层权值为x的边端点并查集合并后， 继续考虑下一层权值的边。
最后
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