HDU 5438 Ponds

Ponds

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 282 Accepted Submission(s): 86


[align=left]Problem Description[/align]
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

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

1
7 7

1 2 3 4 5 6 7

1 4

1 5

4 5

2 3

2 6

3 6

2 7

[align=left]Sample Output[/align]

21

[align=left]Source[/align]
2015 ACM/ICPC Asia Regional Changchun Online
题目大意就是 给出每个点的价值和图 然后删除依次删除读数小于2的顶点，最后剩下的奇数点的联通分量的价值和。
题目的意思第一次没看懂，wa了好几发.
比如这组数据
1
8 8
1 1 1 1 1 1 1 1
1 2
2 3
3 1
4 5
5 6
6 7
7 4
7 8

答案是 3
拓扑排序的思想+dfs判断奇数顶点数的联通块

/* ***********************************************
Author        :pk28
Created Time  :2015/9/13 19:18:07
File Name     :4.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10005
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
return a>b;
}
int n,m;
struct node{
int v,next;

}edge[maxn*20];

int l,pre[maxn];
int w[maxn];
int du[maxn];
int vis[maxn];

void add(int u,int v){
edge[l].v=v;
edge[l].next=pre[u];
pre[u]=l++;
}
queue<int>q;
void init(){
memset(pre,-1,sizeof pre);
cle(du);
l=0;
cle(vis);
cle(w);
while(!q.empty())q.pop();
}
ll sum,ans,cnt;
void dfs(int u){
for(int i=pre[u];i+1;i=edge[i].next){
int v=edge[i].v;
if(!vis[v]){
vis[v]=1;
cnt++;
sum+=w[v];
dfs(v);
}
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int t,x,y;
cin>>t;
while(t--){
init();
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++)scanf("%d",&w[i]);
for(int i=1;i<=m;i++){
scanf("%d%d",&x,&y);
add(x,y);
add(y,x);
du[x]++;
du[y]++;
}
for(int i=1;i<=n;i++){
if(du[i]<2){
q.push(i);
}
}
while(!q.empty()){
int tmp=q.front();
vis[tmp]=1;
q.pop();
for(int i=pre[tmp];i+1;i=edge[i].next){
int v=edge[i].v;
du[v]--;
if(du[v]<2&&!vis[v])q.push(v);
}
}

sum=0;cnt=0;
ans=0;
for(int i=1;i<=n;i++){
if(!vis[i]){
dfs(i);
if(cnt&1) ans+=sum;
cnt=0;
sum=0;
}
}
printf("%I64d\n",ans);
}
return 0;
}