Arpa's weak amphitheater and Mehrdad's valuable Hoses CodeForces - 742D

Arpa’s weak amphitheater and Mehrdad’s valuable Hoses CodeForces - 742D

题目描述
Just to remind, girls in Arpa’s land are really nice.

Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, …, ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.

Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa’s amphitheater can hold at most w weight on it.

Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn’t exceed w.

Input

The first line contains integers n, m and w (1  ≤  n  ≤  1000, , 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.

The second line contains n integers w1, w2, …, wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.

The third line contains n integers b1, b2, …, bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.

The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.

Output

Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn’t exceed w.

Example

Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
大致题意：有n个人，每个人都有颜值bi与体重wi。剧场的容量为W。n个人之间有m种关系表示两个人x，y为好朋友，好朋友的好朋友也是好朋友，她们会形成一个小团体。 每个小团体要么全部参加舞会，要么只参加一个人。保证总重量不超过W，剧场中的颜值最大能到多少

思路
并查集+背包，先用并查集找到每个小团体，然后就是01背包问题了

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
typedef long long ll;
using namespace std;
int tree[1005],dp[1005];

vector<int>group[1005];

int max(int x,int y)
{
    return x>y?x:y;
}
int find(int x)
{
    if(tree[x]==x)
      return x;
    return tree[x]=find(tree[x]); 

}
void megre(int x,int y)
{
    int x1=find(x);
    int y1=find(y);
    if(x1!=y1)
    tree[x1]=y1;
}
int main()
{
     int n,m,wi,sw,sb;
     int w[1005],b[1005],x,y;
      cin>>n>>m>>wi;
     for(int i=1;i<=n;i++)
     cin>>w[i];
     for(int i=1;i<=n;i++)
     cin>>b[i];
     for(int i=1;i<=n;i++)
     tree[i]=i;
     for(int i=1;i<=m;i++)
     {

        cin>>x>>y;
        megre(x,y);
     }
    for(int i=0;i<1005;i++)
    group[i].clear();
    for(int i=1;i<=n;i++)
    group[find(i)].push_back(i);
     memset(dp,0,sizeof(dp));
     for(int i=1;i<=n;i++)
     {
        if(find(i)!=i)
        continue;
        for(int j=wi;j>=0;j--)
        {
            sb=sw=0;
            for(int k=0;k<group[i].size();k++)
            {
                int num=group[i][k];
                sw+=w[num];
                sb+=b[num];
                if(j>=w[num])
                dp[j]=max(dp[j],dp[j-w[num]]+b[num]);//考虑只去一个人的情况
             }
             if(j>=sw)
             dp[j]=max(dp[j],dp[j-sw]+sb);
         }
     }
    cout<<dp[wi];
    return 0;
}