原题链接
考察:拓扑排序+并查集
错误思路:
  离线处理,\(d[i]\)记录i的入度.如果\(d[i]<=1\)就不纳入并查集,否则就加入.
错误原因:
  删除一个点,可能使别的点\(d[i]<=1\)
思路:
  因为\(d[i]\)是会级联影响的,所以我们用拓扑排序求\(d[i]<=1\)的点.但是注意题目是无向边,我们要特判环防止TLE.

Code

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 10010,M = 100010;
struct Road{
	int to,ne;
}road[M<<1];
int n,m,d[N],idx,h[N],q[N],p[N],L[N],R[N];
LL w[N];
bool vis[N];
int sz[N];
void add(int a,int b)
{
	road[idx].to =b,road[idx].ne = h[a],h[a] = idx++;
}
int findf(int x)
{
	if(p[x]!=x) return p[x] = findf(p[x]);
	return p[x];
}
void merge(int a,int b)
{
	int pa = findf(a),pb = findf(b);
	if(pa==pb) return;
	sz[pb]+=sz[pa];
	w[pb]+=w[pa];
	p[pa] = pb;
}
void topsort()
{
	int hh = 0,tt=-1;
	for(int i=1;i<=n;i++)
	  if(d[i]<=1) q[++tt] = i;
	while(hh<=tt)
	{
		int u = q[hh++];
		vis[u] = 1;
		for(int i=h[u];~i;i=road[i].ne)
		{
			int v = road[i].to;
			if(vis[v]) continue;
			if(--d[v]<=1) q[++tt] = v;
		}
	}
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		memset(h,-1,sizeof h);
		memset(d,0,sizeof d);
		memset(vis,0,sizeof vis);
		scanf("%d%d",&n,&m);
		idx =0;
		for(int i=1;i<=n;i++) scanf("%lld",&w[i]),p[i] = i,sz[i] =1;
		for(int i=1;i<=m;i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			d[a]++,d[b]++;
			add(a,b); add(b,a);
			L[i] = a,R[i] = b;
		}
		topsort();
		for(int i=1;i<=m;i++)
		{
			if(vis[L[i]]||vis[R[i]]) continue;
			merge(L[i],R[i]);
		}
		LL res = 0;
		for(int i=1;i<=n;i++)
		{
			int pa = findf(i);
			if(vis[pa]) continue;
			vis[pa] = 1;
			if(sz[pa]&1) res+=w[pa];
		}
		printf("%lld\n",res);
	}
	return 0;
}