codevs 3372 选学霸(hash+并查集+多重背包)

先通过并查集处理出来有多少种不同的集合，每一个集合有多少人。一定要不要忘记了与别的没有联系的独立点。

并查集的时候能够通过hash处理出来每一个数目同样的集合的个数。

这样以人数为权值。个数为限制进行多重背包，结果就是答案。

题目链接：http://codevs.cn/problem/3372/

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define mod 1000000007

const int maxn = 30010;

using namespace std;

int vis1[maxn];
int vis2[maxn];
int vis[maxn];
int num[maxn];

int dp[2*maxn];
int fa[maxn];

struct node
{
    int snum;
    int sum;
} p[maxn];

int n;
void init()
{
    for(int i = 0; i <= n; i++) fa[i] = i;
    memset(vis1, 0, sizeof(vis1));
    memset(vis2, 0, sizeof(vis2));
    memset(dp, 0, sizeof(dp));
    memset(vis, 0, sizeof(vis));
}

int Find(int x)
{
    if(x != fa[x]) fa[x] = Find(fa[x]);
    return fa[x];
}



void add(int x, int y)
{
    int x1, y1;
    x1 = Find(x);
    y1 = Find(y);
    if(x1 != y1) fa[x1] = y1;
}

int main()
{
    int m, k;
    while(~scanf("%d %d %d",&n, &m, &k))
    {
        init();
        int x, y;
        for(int i = 0; i < k; i++)
        {
            scanf("%d %d",&x, &y);
            vis[x] = 1;
            vis[y] = 1;
            add(x, y);
        }
        int ans = 0;
        int xsum = 0;
        for(int i = 1; i <= n; i++)
        {
            if(!vis[i])
            {
                xsum ++;
                continue;
            }
            int s = Find(i);
            vis1[s] ++;
        }
        for(int i = 1; i <= n; i++)
        {
            if(!vis1[i]) continue;
            num[ans++] = vis1[i];
        }
        sort(num, num+ans);
        for(int i = 0; i < ans; i++) vis2[num[i]]++;
        int cnt = 0;
        for(int i = 1; i <= n; i++)
        {
            if(!vis2[i]) continue;
            p[cnt].snum = i;
            p[cnt++].sum = vis2[i];
        }
        int v = 2*(m+1);
        dp[0] = 1;
        for(int i = 0; i < cnt; i++)
        {
            for(int j = v; j >= 0; j--)
            {
                if(!dp[j]) continue;
                for(int kk = 1; kk <= p[i].sum; kk++)
                {
                    if(kk*p[i].snum+j > v) break;
                    if(dp[kk*p[i].snum+j]) break;
                    dp[kk*p[i].snum+j] = 1;
                }
            }
        }
        int lx, rx;
        for(int i = m; i >= 0; i--)
        {
            if(dp[i])
            {
                lx = i;
                break;
            }
        }
        for(int i = m; i <= 2*(m+1); i++)
        {
            if(dp[i])
            {
                rx = i;
                break;
            }
        }
        lx = max(lx, 0);
        rx = min(rx, 2*(m+1));
        int sx = abs(m-lx);
        int sy = abs(rx-m);
        if(sx <= xsum)
        {
            sx = 0;
            lx = m;
        }
        else
        {
            sx -= xsum;
            lx += xsum;
        }
        if(sy < sx)
        {
            cout<<rx<<endl;
            continue;
        }
        cout<<lx<<endl;
    }
    return 0;
}
/*
10 4 9
8 2
1 5
5 10
9 7
10 3
3 4
4 6
8 9
6 8
0

5 3 3
1 2
2 3
3 4
4

*/