Practice link:https://vjudge.net/problem/HDU-6763

題目描述

Problem Description

There arencities andmbidirectional roads in Byteland. These cities are labeled by1,2,…,n, the brightness of thei-th city isbi.

Magician Sunset wants to play a joke on Byteland by making a total eclipse such that the brightness of every city becomes zero. Sunset can do the following operations for arbitrary number of times:

· Select an integerk(1≤k≤n).

· Selectkdistinct citiesc1,c2,…,ck(1≤ci≤n) such that they are connected with each other. In other words, for every pair of distinct selected citiesciandcj(1≤i<j≤k), if you are at cityci, you can reach citycjwithout visiting cities not in{c1,c2,…,ck}

.

· For every selected cityci(1≤i≤k), decreasebciby1.

Note that Sunset will always choosekwith the maximum possible value. Now Sunset is wondering what is the minimum number of operations he needs to do, please write a program to help him.

Input

The first line of the input contains a single integerT

(1≤T≤10), the number of test cases.

For each case, the first line of the input contains two integersnandm(1≤n≤100000,1≤m≤200000), denoting the number of cities and the number of roads.

The second line of the input containsnintegersb1,b2,…,bn(1≤bi≤109), denoting the brightness of each city.

Each of the followingmlines contains two integersuiandvi(1≤ui,vi≤n,ui≠vi), denoting an bidirectional road between theui-th city and thevi-th city. Note that there may be multiple roads between the same pair of cities.

Output

For each test case, output a single line containing an integer, the minimum number of operations.

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 #define MOD 998244353 
 4 #define INF 0x3f3f3f3f
 5 #define mem(a,x) memset(a,x,sizeof(a))  
 6 #define _for(i,a,b) for(int i=a; i< b; i++)
 7 #define _rep(i,a,b) for(int i=a; i<=b; i++)
 8 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 9 using namespace std;
10 const int maxn = 100000;
11 int fa[maxn+5];
12 int vis[maxn+5];
13 int a[maxn+5];
14 vector<int>g[maxn+5];
15 int n,m;
16 ll ans;
17 struct IN{
18     int id,w;
19 }ss[maxn+5];
20 bool cmp(IN x,IN y){
21     return x.w>y.w;
22 }
23 int find(int x)
24 {
25     return (fa[x]==x)?x:(fa[x]=find(fa[x]));
26 }
27 void merge(int i,int j)
28 {
29     int x=i;
30     int y=find(j);
31     fa[y]=x;
32 }
33 int main()
34 {
35     int t;
36     scanf("%d",&t);
37     while(t--){
38         ans=0;
39         scanf("%d %d",&n,&m);
40         for(int i=1;i<=n;i++){
41             scanf("%d",&ss[i].w);
42             fa[i]=i;
43             a[i]=ss[i].w;
44             ss[i].id=i;
45             g[i].clear();
46             vis[i]=0;
47         }
48         for(int i=1;i<=m;i++){
49             int u,v;
50             scanf("%d %d",&u,&v);
51             g[u].push_back(v);
52             g[v].push_back(u);
53         }
54         sort(ss+1,ss+n+1,cmp);
55         for(int j=1;j<=n;j++){
56             int now=ss[j].id;
57             ans+=a[now];
58             vis[now]=1;
59             for(int i=0;i<g[now].size();i++){
60                 int v=g[now][i];
61                 if(!vis[v]||find(v)==find(now))continue;
62                 merge(v,now);
63                 ans-=a[now];
64             }
65         }
66         cout<<ans<<endl;
67     }
68     return 0;
69 }
70 
71 /*8 5
72   1 3 6 5 4 2 1 1
73   1 7
74   3 1
75   6 3
76   5 8
77   2 4   */