2015长春网络赛 —— B. Ponds （拓扑排序删点+DFS）

题目描述

Description
Betty owns a lot of ponds, some of them are connected with other ponds by pipes, and there will not be more than one pipe between two ponds. Each pond has a value v.
Now Betty wants to remove some ponds because she does not have enough money. But each time when she removes a pond, she can only remove the ponds which are connected with less than two ponds, or the pond will explode.
Note that Betty should keep removing ponds until no more ponds can be removed. After that, please help her calculate the sum of the value for each connected component consisting of a odd number of ponds

Input
The first line of input will contain a number T(1≤T≤30) which is the number of test cases.
For each test case, the first line contains two number separated by a blank. One is the number p(1≤p≤104) which represents the number of ponds she owns, and the other is the number m(1≤m≤105) which represents the number of pipes.
The next line contains p numbers v1,…,vp, where vi(1≤vi≤108) indicating the value of pond i.
Each of the last m lines contain two numbers a and b, which indicates that pond a and pond b are connected by a pipe.

Output
For each test case, output the sum of the value of all connected components consisting of odd number of ponds after removing all the ponds connected with less than two pipes.

Samples
Input Copy
1
7 7
1 2 3 4 5 6 7
1 4
1 5
4 5
2 3
2 6
3 6
2 7
Output
21
Source
2015 ACM/ICPC Asia Regional Changchun Online

题意：

n点m边无向图，问将度数<2的点删除后，剩余的由奇数个点构成的连通块的权值之和。

思路：

拓扑排序删点，然后DFS。

代码：

#pragma GCC optimize(3)
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline")
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll>PLL;
typedef pair<int,int>PII;
typedef pair<double,double>PDD;
#define I_int ll
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
char F[200];
inline void out(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
    //cout<<" ";
}
ll ksm(ll a,ll b,ll p){ll res=1;while(b){if(b&1)res=res*a%p;a=a*a%p;b>>=1;}return res;}
const int inf=0x3f3f3f3f,mod=1e9+7;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int maxn=1e5+7,maxm=3e5+7,N=1e6+7;
const double PI = atan(1.0)*4;

int h[N],e[N],ne[N],idx,ans=N,n,m;
int d[N];
int val[maxn];
queue<int>q;
bool vis[maxn];

void init(){
    idx=0;
    memset(h,-1,sizeof h);
    memset(vis,0,sizeof vis);
    memset(d,0,sizeof d);
}

void add(int a,int b){
    e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}

void topsort(){
    for(int i=1;i<=n;i++)
        if(d[i]<2) q.push(i),vis[i]=1;
    while(!q.empty()){
        int t=q.front();q.pop();
        for(int i=h[t];~i;i=ne[i]){
            int j=e[i];
            if(vis[j]) continue;
            if(--d[j]<2) q.push(j),vis[j]=1;
        }
    }
}

ll siz=0,sum=0;

void dfs(int u){
    siz++;
    vis[u]=1;
    sum+=val[u];
    for(int i=h[u];~i;i=ne[i]){
        int j=e[i];
        if(vis[j]) continue;
        dfs(j);
    }
}

int main(){
    int T=read();
    while(T--){
        init();
        n=read(),m=read();
        for(int i=1;i<=n;i++) val[i]=read();
        while(m--){
            int u=read(),v=read();
            add(u,v);add(v,u);
            d[u]++;d[v]++;
        }
        topsort();
        ll res=0;
        siz=sum=0;
        for(int i=1;i<=n;i++)
            if(!vis[i]){
                siz=sum=0;
                dfs(i);
                if(siz%2) res+=sum;
            }
        out(res);puts("");
        while(!q.empty()) q.pop();
    }
    return 0;
}