Marriage Match III HDU - 3277（二分权值 + 拆点 建边）

Posted 2021-01-14 wtsruvf

题意：

　　只不过是hdu3081多加了k种选择

　　想一下，最多能玩x轮，是不是就是每个女生能最多选x个男生

　　现在题中的每个女生比3081多了k中选择   那就把女生拆点  i  i‘  

　　i --> i‘ 连一条权值为K的边  如果男女无争吵 连上i --> 男  权值为1 

　　有争吵 连上i‘ --> 男 权值为1

　　

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define pd(a) printf("%d
", a);
#define plld(a) printf("%lld
", a);
#define pc(a) printf("%c
", a);
#define ps(a) printf("%s
", a);
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 100100, INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
int n, m, z, s, t, cnt, p;
int f[maxn], re[300][300];
int d[maxn], cur[maxn], head[maxn];
struct node
{
    int u, v, c, next;
}Node[maxn << 1];

void add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    Node[cnt].next = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c);
    add_(v, u, 0);
}

bool bfs()
{
    queue<int> Q;
    mem(d, 0);
    Q.push(s);
    d[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = head[u]; i != -1; i = Node[i].next)
        {
            node e = Node[i];
            if(!d[e.v] && e.c > 0)
            {
                d[e.v] = d[u] + 1;
                Q.push(e.v);
                if(e.v == t) return 1;
            }
        }
    }
    return d[t] != 0;
}

int dfs(int u, int cap)
{
    int ret = 0;
    if(u == t || cap == 0)
        return cap;
    for(int &i = cur[u]; i != -1; i = Node[i].next)
    {
        node e = Node[i];
        if(d[e.v] == d[u] + 1 && e.c > 0)
        {
            int V = dfs(e.v, min(cap, e.c));
            Node[i].c -= V;
            Node[i ^ 1].c += V;
            ret += V;
            cap -= V;
            if(cap == 0) break;
        }
    }
    if(cap > 0) d[u] = -1;
    return ret;
}

int Dinic()
{
    int ans = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof(head));
        ans += dfs(s, INF);
    }
    return ans;
}

int find(int x)
{
    return f[x] == x ? x : (f[x] = find(f[x]));
}

void build(int mid)
{
    for(int i = 1; i <= n; i++)
    {
        add(s, i, mid);
        add(n + i, t, mid);
        add(i, n * 2 + i, p);
    }
    for(int i = 1; i <= n; i++)
        for(int k = n + 1; k <= n * 2; k++)
        {
            if(re[i][k]) add(i, k, 1);
            else add(n * 2 + i, k, 1);
        }
}

void init()
{
    for(int i = 1; i < maxn; i++) f[i] = i;
    mem(re, 0);
    mem(head, -1);
    cnt = 0;
}

int main()
{
    int T;
    rd(T);
    while(T--)
    {
        init();
        int a, b;
        rd(n), rd(m), rd(p), rd(z);
        s = 0, t = n * 3 + 1;
        for(int i = 0; i < m; i++)
        {
            rd(a), rd(b);
            b += n;
            re[a][b] = re[b][a] = 1;
        }
        for(int i = 0; i < z; i++)
        {
            rd(a), rd(b);
            int r = find(a);
            int l = find(b);
            if(r != l) f[l] = r;
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = i + 1; j <= n; j++)
            {
                if(find(i) == find(j))
                    for(int k = n + 1; k <= n * 2; k++)
                        re[i][k] = re[j][k] = (re[i][k] || re[j][k]);
            }
        }
        int x = 0, y = 300;
        while(x <= y)
        {
            mem(head, -1);
            cnt = 0;
            int mid = x + (y - x) / 2;
            build(mid);
            if(Dinic() == mid * n) x = mid + 1;
            else y = mid - 1;
        }
        pd(y);

    }

    return 0;
}