hdu3081 二分+并查集+最大流

http://acm.hdu.edu.cn/showproblem.php?pid=3081
题意：
有n个女孩，n个男孩，对于每个女孩有几个没有争吵过的男孩，每个女孩有几个朋友（都是女孩），对于没有争吵过的男孩可以建立关系。这样就可以进行一次游戏，那么下一次可以选择一个之前没有选择过的人建立关系。那么求最多能进行多少次关系。

思路：
对于朋友关系，很容易想到用并查集维护，那么在一个集合中所连的边都是一样的。二分答案，判断mid轮的可行性。将源点连到每个女孩，容量为mid。将女孩连到相应的男孩上，容量为1。将每个男孩连到汇点，容量为mid。如果mid轮能满足的话，那么提高下限，不行就缩小上限。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <cmath>using namespace std;const int M_node = 400,M_edge = 20009000, INF = 0x3f3f3f3f;typedef pair<int,int> pii;struct edge{    int to,cap,next;}edge[M_edge];bool mp[M_node][M_node];pii in[20009];int p[M_node];int head[M_node],level[M_node];int n,m,f,num,numuni;int s,t;void init(){    for(int i = 0;i <= n;i++)    {        //V[i].clear();        //V1[i].clear();        p[i] = i;        //nump[i] = 1;    }    //memset(mp,false,sizeof(mp));    //numuni = 1;}void add_edge(int u,int v,int cap){    edge[num].to = v;edge[num].cap = cap;edge[num].next = head[u];head[u] = num++;    edge[num].to = u;edge[num].cap = 0;edge[num].next = head[v];head[v] = num++;}int find(int x){    return p[x] == x ? x : p[x] = find(p[x]);}void uni(int x,int y){    x = find(x);    y = find(y);    if(x != y)    {        p[x] = y;        //nump[y] += nump[x];    }}void build(int mid){    memset(head,-1,sizeof(head));    memset(mp,false,sizeof(mp));    num = 0;    for(int i = 1;i <= n;i++)    {        add_edge(s,i,mid);        add_edge(i+n,t,mid);    }    for(int i = 0;i < m;i++)    {        pii tmp = in[i];        int u = tmp.first;        int v = tmp.second;        for(int j = 1;j <= n;j++)        {            if(find(u) == find(j) && !mp[j][v])            {                mp[j][v] = true;                add_edge(j,v+n,1);            }        }    }}bool bfs(int s,int t){    memset(level,-1,sizeof(level));    level[s] = 0;    queue<int> q;    q.push(s);    while(!q.empty())    {        int v = q.front();        q.pop();        for(int i = head[v];i != -1;i = edge[i].next)        {            int u = edge[i].to;            if(level[u] == -1 && edge[i].cap > 0)            {                level[u] = level[v] + 1;                q.push(u);            }        }    }    if(level[t] != -1) return true;    return false;}int dfs(int v,int t,int f){    if(v == t) return f;    for(int i = head[v];i != -1 ;i = edge[i].next)    {        int u = edge[i].to;        if(level[u] > level[v] && edge[i].cap > 0)        {            int d = dfs(edge[i].to,t,min(f,edge[i].cap));            if(d > 0)            {                edge[i].cap -= d;                edge[i^1].cap += d;                return d;            }        }    }    level[v] = -1; //优化    return 0;}int dinic(int s,int t){    int flow = 0;    while(bfs(s,t))    {        int f = 0;        while((f = dfs(s,t,INF)) > 0) flow += f;    }    return flow;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&f);        init();        for(int i = 0;i < m;i++)        {            int a,b;            scanf("%d %d",&a,&b);            in[i] = make_pair(a,b);        }        for(int i = 0;i < f;i++)        {            int a,b;            scanf("%d%d",&a,&b);            uni(a,b);        }        s = 0;        t = 2*n + 1;        int l = -1, r = n+1;        while(r - l > 1)        {            int mid = (l+r)>>1;            build(mid);            int flow = dinic(s,t);            if(flow >= n*mid) l = mid;            else r = mid;        }        printf("%d\n",l);    }    return 0;}