hdu 3277(二分+最大流+拆点+离线处理+模板问题...)

Marriage Match III

Time Limit: 10000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1971 Accepted Submission(s): 583


[align=left]Problem Description[/align]
Presumably,
you all have known the question of stable marriage match. A girl will
choose a boy; it is similar as the ever game of play-house . What a
happy time as so many friends play together. And it is normal that a
fight or a quarrel breaks out, but we will still play together after
that, because we are kids.

Now, there are 2n kids, n boys
numbered from 1 to n, and n girls numbered from 1 to n. As you know,
ladies first. So, every girl can choose a boy first, with whom she has
not quarreled, to make up a family. Besides, the girl X can also choose
boy Z to be her boyfriend when her friend, girl Y has not quarreled with
him. Furthermore, the friendship is mutual, which means a and c are
friends provided that a and b are friends and b and c are friend.

Once
every girl finds their boyfriends they will start a new round of this
game—marriage match. At the end of each round, every girl will start to
find a new boyfriend, who she has not chosen before. So the game goes on
and on. On the other hand, in order to play more times of marriage
match, every girl can accept any K boys. If a girl chooses a boy, the
boy must accept her unconditionally whether they had quarreled before or
not.

Now, here is the question for you, how many rounds can these 2n kids totally play this game?

[align=left]Input[/align]
There are several test cases. First is an integer T, means the number of test cases.
Each
test case starts with three integer n, m, K and f in a line
(3<=n<=250, 0<m<n*n, 0<=f<n). n means there are 2*n
children, n girls(number from 1 to n) and n boys(number from 1 to n).
Then m lines follow. Each line contains two numbers a and b, means girl a and boy b had never quarreled with each other.
Then f lines follow. Each line contains two numbers c and d, means girl c and girl d are good friends.

[align=left]Output[/align]
For each case, output a number in one line. The maximal number of Marriage Match the children can play.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

3

[align=left]Author[/align]
starvae

这个题的题意hdu 3081 几乎一样,多了个条件,就是女生还可以任选k个自己不喜欢的男生,所以等于女生多了一个容量限制,所以不能够用二分图解了，将女生拆点,将女生 i 和女生 n+i 之间连一条容量为k的边,i女生和喜欢的男生连容量为1的边,i+n和不喜欢的男生连容量为1的边,然后二分枚举完美配对上限，但是这题还卡时间,所以我们要将输入的男女生进行离线处理，先将所有的祖先结点和男生配对，然后再去寻找其孩子结点进行配对,这样的话就可以省掉一层循环了。自己的Dinic模板过不了,在网上找了个大神模板。不想学ISAP。。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXNODE = 805;
const int MAXEDGE = MAXNODE*MAXNODE;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge
{
int u, v;
Type cap, flow;
Edge() {}
Edge(int u, int v, Type cap, Type flow)
{
this->u = u;
this->v = v;
this->cap = cap;
this->flow = flow;
}
};

struct Dinic
{
int n, m, s, t;
Edge edges[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool vis[MAXNODE];
Type d[MAXNODE];
int cur[MAXNODE];
vector<int> cut;

void init(int n)
{
this->n = n;
memset(first, -1, sizeof(first));
m = 0;
}
void add_Edge(int u, int v, Type cap)
{
edges[m] = Edge(u, v, cap, 0);
next[m] = first[u];
first[u] = m++;
edges[m] = Edge(v, u, 0, 0);
next[m] = first[v];
first[v] = m++;
}

bool bfs()
{
memset(vis, false, sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = true;
while (!Q.empty())
{
int u = Q.front();
Q.pop();
for (int i = first[u]; i != -1; i = next[i])
{
Edge& e = edges[i];
if (!vis[e.v] && e.cap > e.flow)
{
vis[e.v] = true;
d[e.v] = d[u] + 1;
Q.push(e.v);
}
}
}
return vis[t];
}

Type dfs(int u, Type a)
{
if (u == t || a == 0) return a;
Type flow = 0, f;
for (int &i = cur[u]; i != -1; i = next[i])
{
Edge& e = edges[i];
if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0)
{
e.flow += f;
edges[i^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}

Type Maxflow(int s, int t)
{
this->s = s;
this->t = t;
Type flow = 0;
while (bfs())
{
for (int i = 0; i < n; i++)
cur[i] = first[i];
flow += dfs(s, INF);
}
return flow;
}
void MinCut()
{
cut.clear();
for (int i = 0; i < m; i += 2)
{
if (vis[edges[i].u] && !vis[edges[i].v])
cut.push_back(i);
}
}
} gao;
const int N = 251;
bool vis

;
int father
;
int girl[N*N],boy[N*N]; ///这里m是属于 (0,n*n]的
int n,m,k,f,src,des;
int _find(int x)
{
if(father[x]!=x)
{
father[x] = _find(father[x]);
}
return father[x];
}
void build(int c)
{
gao.init(3*n+2);
for(int i=1; i<=n; i++)
{
gao.add_Edge(src,i,c);
gao.add_Edge(i,n+i,k);
gao.add_Edge(2*n+i,des,c);
}
for(int i=1; i<=n; i++)
{
for(int j=2*n+1; j<=3*n; j++)
{
if(vis[i][j]) gao.add_Edge(i,j,1);
else gao.add_Edge(n+i,j,1);
}
}
}
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--)
{
scanf("%d%d%d%d",&n,&m,&k,&f);
memset(vis,false,sizeof(vis));
src = 0,des = 3*n+1;
for(int i=1; i<=n; i++) father[i] = i;
/* for(int i=1;i<=m;i++){  //TLE
int u,v;
scanf("%d%d",&u,&v);
v+=2*n;
vis[u][v] = true;
}*/
for(int i=1; i<=m; i++)
{
scanf("%d%d",&girl[i],&boy[i]);
boy[i]+=2*n;
}
for(int i=1; i<=f; i++)
{
int u,v;
scanf("%d%d",&u,&v);
int a = _find(u),b = _find(v);
if(a!=b)
father[a] = b;
}
for(int i=1; i<=m; i++)
{
vis[_find(girl[i])][boy[i]] = true;
}
for(int i=1; i<=n; i++) ///预处理所有关系
{
for(int j=2*n+1; j<=3*n; j++)
{
if(vis[i][j]) continue;
if(vis[_find(i)][j]) vis[i][j] = true;
}
}
/*for(int i=1;i<=n;i++){ //TLE
for(int j=1;j<=n;j++){
if(_find(i)==_find(j)){
for(int k=2*n+1;k<=3*n;k++){
if(vis[i][k]||vis[j][k]) vis[i][k] = vis[j][k] = 1;
}
}
}
}*/
int l = 0,r = n,ans = 0;
while(l<=r)
{
int mid = (l+r)>>1;
build(mid);
if(gao.Maxflow(src,des)==mid*n)
{
ans = mid;
l = mid+1;
}
else r = mid-1;
}
printf("%d\n",ans);
}
return 0;
}