解题报告 之 POJ2289 Jamie‘s Contact Groups

Description

Jamie is a very popular girl and has quite a lot of friends, so she always keeps a very long contact list in her cell phone. The contact list has become so long that it often takes a long time for her to browse through the whole list to find a friend‘s number.
As Jamie‘s best friend and a programming genius, you suggest that she group the contact list and minimize the size of the largest group, so that it will be easier for her to search for a friend‘s number among the groups. Jamie takes your advice and gives you
her entire contact list containing her friends‘ names, the number of groups she wishes to have and what groups every friend could belong to. Your task is to write a program that takes the list and organizes it into groups such that each friend appears in only
one of those groups and the size of the largest group is minimized.

Input

There will be at most 20 test cases. Ease case starts with a line containing two integers N and M. where N is the length of the contact list and M is the number of groups. N lines then follow. Each line contains a friend‘s name and the groups the friend could
belong to. You can assume N is no more than 1000 and M is no more than 500. The names will contain alphabet letters only and will be no longer than 15 characters. No two friends have the same name. The group label is an integer between 0 and M - 1. After the
last test case, there is a single line `0 0‘ that terminates the input.

Output

For each test case, output a line containing a single integer, the size of the largest contact group.

Sample Input

3 2
John 0 1
Rose 1
Mary 1
5 4
ACM 1 2 3
ICPC 0 1
Asian 0 2 3
Regional 1 2
ShangHai 0 2
0 0

Sample Output

2
2

题目大意：有n个联系人，m个分组，告诉你每个联系人可以分到哪些分组，且每个联系人只需要分到一个分组即可。现在问所有组的人数中最大的那一个的最小值是多少？

分析：最大值最小用二分来解决，然后我们如何考虑如何在图中验证是否可以满足当前值mid，答案是用最大流来试验。首先超级源点连接每一个联系人节点，负载为1，然后每个联系人连接所以可以被分入的分组节点，负载为1（因为超级源点限制了所以不用拆点），然后所有分组节点连接到超级汇点，负载为mid，表示所有分组容量不超过mid，然后跑最大流，根据结果二分。

上代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
using namespace std;

const int MAXM = 900000;
const int MAXN = 2000;
const int INF = 0x3f3f3f3f;

struct Edge
{
	int to, cap, next;
};

Edge edge[MAXM];
int level[MAXN];
int head[MAXN];
int group[MAXN][MAXN];
int src, des, cnt;

void addedge( int from, int to, int cap )
{

	edge[cnt].to = to;
	edge[cnt].cap = cap;
	edge[cnt].next = head[from];
	head[from] = cnt++;

	swap( from, to );

	edge[cnt].to = to;
	edge[cnt].cap = 0;
	edge[cnt].next = head[from];
	head[from] = cnt++;
}

int bfs( )
{
	memset( level, -1, sizeof level );
	queue<int>q;
	while (!q.empty( ))
		q.pop( );

	level[src] = 0;
	q.push( src );

	while (!q.empty( ))
	{
		int u = q.front( );
		q.pop( );

		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].to;
			if (edge[i].cap > 0 && level[v] == -1)
			{
				level[v] = level[u] + 1;
				q.push( v );
			}
		}
	}
	return level[des] != -1;
}

int dfs( int u, int f )
{
	if (u == des) return f;
	int tem;
	for (int i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].to;
		if (edge[i].cap>0 && level[v] == level[u] + 1)
		{
			tem = dfs( v, min( f, edge[i].cap ) );
			if (tem > 0)
			{
				edge[i].cap -= tem;
				edge[i ^ 1].cap += tem;
				return tem;
			}
		}
	}
	level[u] = -1;
	return 0;
}

int Dinic( )
{
	int ans = 0, tem;

	while (bfs( ))
	{
		while ((tem = dfs( src, INF )) > 0)
		{
			ans += tem;
		}
	}
	return ans;
}

int main( )
{
	int n, m;
	src = 0;
	des = 1980;
	string str;
	while (cin >> n >> m&&n&&m)
	{
		getchar();
		for (int i = 1; i <= n; i++)
		{
			getline( cin, str );
			stringstream ss(str);
			ss >> str;
			int g;
			int num = 0;
			while (ss >> g)
			{
				group[i][++num] = g;
			}
			group[i][0] = num;
		}

		int low = 0, high = INF - 1;
		int ans = -1;
		while (low <= high)
		{
			int mid = (low + high) / 2;
			memset( head, -1, sizeof head );
			cnt = 0;

			for (int i = 1; i <= n; i++)
			{
				addedge( src, i, 1 );
			}
			for (int i = 1; i <= m; i++)
			{
				addedge( i + 1000, des, mid );
			}

			for (int i = 1; i <= n; i++)
			{
				for (int j = 1; j <= group[i][0]; j++)
				{
					addedge( i, group[i][j] + 1001, 1 );
				}
			}
			if (Dinic() < n) low = mid + 1;
			else
			{
				ans = mid;
				high = mid - 1;
			}

		}
		cout << ans << endl;
	}
	return 0;
}

这年头想要抱个大腿真困难，，还是得靠自个儿！

解题报告 之 POJ2289 Jamie's Contact Groups