Codeforces - E. Tourism

E. Tourism

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Alex decided to go on a touristic trip over the country.

For simplicity let’s assume that the country has n cities and m bidirectional roads connecting them. Alex lives in city s and initially located in it. To compare different cities Alex assigned each city a score wi which is as high as interesting city seems to Alex.

Alex believes that his trip will be interesting only if he will not use any road twice in a row. That is if Alex came to city v from city u, he may choose as the next city in the trip any city connected with v by the road, except for the city u.

Your task is to help Alex plan his city in a way that maximizes total score over all cities he visited. Note that for each city its score is counted at most once, even if Alex been there several times during his trip.

Input
First line of input contains two integers n and m, (1≤n≤2⋅105, 0≤m≤2⋅105) which are numbers of cities and roads in the country.

Second line contains n integers w1,w2,…,wn (0≤wi≤109) which are scores of all cities.

The following m lines contain description of the roads. Each of these m lines contains two integers u and v (1≤u,v≤n) which are cities connected by this road.

It is guaranteed that there is at most one direct road between any two cities, no city is connected to itself by the road and, finally, it is possible to go from any city to any other one using only roads.

The last line contains single integer s (1≤s≤n), which is the number of the initial city.

Output
Output single integer which is the maximum possible sum of scores of visited cities.

Examples
inputCopy
5 7
2 2 8 6 9
1 2
1 3
2 4
3 2
4 5
2 5
1 5
2
outputCopy
27
inputCopy
10 12
1 7 1 9 3 3 6 30 1 10
1 2
1 3
3 5
5 7
2 3
5 4
6 9
4 6
3 7
6 8
9 4
9 10
6
outputCopy
61

---

边双+树形dp

先缩点是很明显的，因为一个边双连通分量当中的是可以随便走的。

之后变成了一棵树，然后有些点是可以回来的，有些不能，取决于块的个数，如果大于等于2就可以回来。

那么我们继续缩点，怎么缩点呢？

因为有些是可以直接回来，所以我们直接把权值给他的父亲不就好了，然后把他的权值情况，更新父亲的size。

之后再一遍dfs，就不用考虑是否回来的问题了，全部当成不能回来，因为回来的已经缩点了。

---

AC代码：

#pragma GCC optimize(2)
#include
#define int long long
using namespace std;
const int N=2e5+10,M=4e5+10;
int n,m,dfn[N],low[N],col[N],sum[N],brige[M],w[N],vis[N],co,cnt,s,dp[N],num[N];
int head[N],nex[M],to[M],tot=1;
stack<int> st;	vector<int> v[N];
inline void add(int a,int b){to[++tot]=b; nex[tot]=head[a]; head[a]=tot;}
void Tarjan(int x){
	dfn[x]=low[x]=++cnt; vis[x]=1;	st.push(x);
	for(int i=head[x];i;i=nex[i]){
		if(brige[i])	continue;	brige[i]=brige[i^1]=1;
		if(!dfn[to[i]]){
			Tarjan(to[i]);	low[x]=min(low[x],low[to[i]]);
		}else if(vis[to[i]]) low[x]=min(low[x],dfn[to[i]]);
	}
	if(low[x]==dfn[x]){
		int u; co++;
		do{
			u=st.top();	vis[u]=0;	st.pop();	col[u]=co; sum[co]+=w[u]; num[co]++;
		}while(u!=x);
	}
}
void dfs(int x,int fa){ 
	for(int i=0;i<v[x].size();i++){
		int to=v[x][i];
		if(to==fa)	continue;
		dfs(to,x);
		if(num[to]>=2){
			sum[x]+=sum[to]; sum[to]=0; num[x]+=num[to];
		}
	}
}
void dfs1(int x,int fa){
	dp[x]=sum[x]; int t=0;
	for(int i=0;i<v[x].size();i++){
		int to=v[x][i];
		if(to==fa)	 continue;
		dfs1(to,x);	t=max(t,dp[to]);
	}
	dp[x]+=t;
}
signed main(){
	cin>>n>>m;
	for(int i=1;i<=n;i++)	scanf("%lld",&w[i]);
	for(int i=1,a,b;i<=m;i++)	scanf("%lld %lld",&a,&b),add(a,b),add(b,a);
	for(int i=1;i<=n;i++)	if(!dfn[i])	Tarjan(i);
	for(int i=1;i<=n;i++){
		for(int j=head[i];j;j=nex[j]){
			if(col[i]!=col[to[j]])	v[col[i]].push_back(col[to[j]]);
		}
	}
	cin>>s;	dfs(col[s],-1); dfs1(col[s],-1);	cout<<dp[col[s]]<<endl;
	return 0;
}