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这个题目是说，有n个女的和男的找伴侣。然后女的具有主动选择权，每个女的可以选自己喜欢的男的，也可以挑选k个不喜欢的男的，做法就是：把女的拆点，u1->u2建立一条容量为k的边。如果遇见喜欢的男生i->j+2*n建一条容量为1的边，否则i+n->j+2*n建一条容量为1的边。最后将源点和女生相连容量为mid，汇点与男生相连容量为mid。枚举mid，看是否会产生满流。

可能姿势不够优美dinic超时了啊，换成SAP快了很多啊、、、

Marriage Match III

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4 5 1 2
1 1
2 3
3 2
4 2
4 4
1 4
2 3

3

#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-12
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3ffffff
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;

const int maxn = 1010;



///int deep[maxn];
int cnt;
int n, m, k;
int S, T;
int fa[maxn];
int vis[maxn][maxn];
int cur[maxn], head[maxn];
int dis[maxn], gap[maxn];
int aug[maxn], pre[maxn];

struct node
{
    int v, w;
    int next;
} f[510000];

struct node1
{
    int l, r;
} pp[50010];

int Find(int x)
{
    if(x != fa[x]) fa[x] = Find(fa[x]);
    return fa[x];
}
void Union(int x,int y)
{
    int xx,yy;
    xx=Find(x);
    yy=Find(y);
    if(xx!=yy)
        fa[xx]=yy;
}

void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
    for(int i = 1; i <= n; i++) fa[i] = i;
}
void add(int u, int v, int w)
{
    f[cnt].v = v;
    f[cnt].w = w;
    f[cnt].next = head[u];
    head[u] = cnt++;

    f[cnt].v = u;
    f[cnt].w = 0;
    f[cnt].next = head[v];
    head[v] = cnt++;
}

int SAP(int s, int e, int n)
{
    int max_flow = 0, v, u = s;
    int id, mindis;
    aug[s] = INF;
    pre[s] = -1;
    memset(dis, 0, sizeof(dis));
    memset(gap, 0, sizeof(gap));
    gap[0] = n;
    for (int i = 0; i <= n; ++i)  cur[i] = head[i];/// 初始化当前弧为第一条弧
    while (dis[s] < n)
    {
        bool flag = false;
        if (u == e)
        {
            max_flow += aug[e];
            for (v = pre[e]; v != -1; v = pre[v]) /// 路径回溯更新残留网络
            {
                id = cur[v];
                f[id].w -= aug[e];
                f[id^1].w += aug[e];
                aug[v] -= aug[e]; /// 修改可增广量，以后会用到
                if (f[id].w == 0) u = v; /// 不回退到源点，仅回退到容量为0的弧的弧尾
            }
        }
        for (id = cur[u]; id != -1; id = f[id].next)/// 从当前弧开始查找允许弧
        {
            v = f[id].v;
            if (f[id].w > 0 && dis[u] == dis[v] + 1) /// 找到允许弧
            {
                flag = true;
                pre[v] = u;
                cur[u] = id;
                aug[v] = min(aug[u], f[id].w);
                u = v;
                break;
            }
        }
        if (flag == false)
        {
            if (--gap[dis[u]] == 0) break; ///gap优化，层次树出现断层则结束算法
            mindis = n;
            cur[u] = head[u];
            for (id = head[u]; id != -1; id = f[id].next)
            {
                v = f[id].v;
                if (f[id].w > 0 && dis[v] < mindis)
                {
                    mindis = dis[v];
                    cur[u] = id; /// 修改标号的同时修改当前弧
                }
            }
            dis[u] = mindis + 1;
            gap[dis[u]]++;
            if (u != s) u = pre[u]; /// 回溯继续寻找允许弧
        }
    }
    return max_flow;
}

void build(int mid)
{
    int a, b;
    cnt = 0;
    memset(head, -1, sizeof(head));
    S = 0;
    T = 3*n+1;
    for(int i = 1 ; i <= n; i++)
    {
        add(S, i, mid);  
        add(i+2*n, T, mid);
        add(i, i+n, k);
    }
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < m; i++)
    {
        a = pp[i].l;
        b = pp[i].r;
        vis[Find(a)][b] = 1;
    }
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            if(!vis[Find(i)][j]) add(i+n, j+2*n, 1);
            else add(i, j+2*n, 1);
        }
    }
}
int main()
{
    int K;
    cin >>K;
    while(K--)
    {
        int p;
        scanf("%d %d %d %d",&n, &m, &k, &p);
        init();
        for(int i = 0; i < m; i++) scanf("%d %d",&pp[i].l, &pp[i].r);
        int x, y;
        while(p--)
        {
            scanf("%d %d",&x, &y);
            Union(x, y);
        }
        int l = 0;
        int r = n;
        int ans = 0;
        int mid;
        while(l <= r)
        {
            mid = (l+r)>>1;
            build(mid);
            if(SAP(0, 3*n+1, 3*n+2) >= n*mid)
            {
                l = mid+1;
                ans = mid;
                continue;
            }
            r = mid-1;
        }
        printf("%d\n",ans);
    }
    return 0;
}

HDU 3277 Marriage Match III(拆点+二分+最大流SAP),布布扣,bubuko.com

HDU 3277 Marriage Match III(拆点+二分+最大流SAP)

标签：des   style   blog   color   java   os   io   strong   

原文地址：http://blog.csdn.net/xu12110501127/article/details/38580329