poj 1112 Team Them Up!（建图+dp）

http://poj.org/problem?id=1112

Team Them Up!

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6368		Accepted: 1706		Special Judge

Description

Your task is to divide a number of persons into two teams, in such a way, that: 

everyone belongs to one of the teams; 

every team has at least one member; 

every person in the team knows every other person in his team; 

teams are as close in their sizes as possible. 

This task may have many solutions. You are to find and output any solution, or to report that the solution does not exist.

Input

For simplicity, all persons are assigned a unique integer identifier from 1 to N. 

The first line in the input file contains a single integer number N (2 <= N <= 100) - the total number of persons to divide into teams, followed by N lines - one line per person in ascending order of their identifiers. Each line contains the list of distinct numbers Aij (1 <= Aij <= N, Aij != i) separated by spaces. The list represents identifiers of persons that ith person knows. The list is terminated by 0.

Output

If the solution to the problem does not exist, then write a single message "No solution" (without quotes) to the output file. Otherwise write a solution on two lines. On the first line of the output file write the number of persons in the first team, followed by the identifiers of persons in the first team, placing one space before each identifier. On the second line describe the second team in the same way. You may write teams and identifiers of persons in a team in any order.

Sample Input

5
2 3 5 0
1 4 5 3 0
1 2 5 0
1 2 3 0
4 3 2 1 0

Sample Output

3 1 3 5
2 2 4

Source

Northeastern Europe 2001

题意：N个人，将N个人分成两组，组内的人都相互认识，让两个组的人数尽量接近。

让组内的人都要相互认识，其实就是一个团。所以用求团的思路建图，如果两个人不互相认识，就建一条边。显然如果建出来的图不是二分图，那么一定无解。仔细分析，对于二分图的每一个联通块的人，在二分图不同边的一定不能放在一个组。所以我们求出每一个联通块在二分图不同边的数的数目，用x和y表示。那么对应每一个联通块都有对应的x和y值。那么我们可以用dp求出两个组的人数最接近为多少。在dp的过程中记录路径就可以输出方案。代码如下：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<vector>
#define nn 110
#define inff 0x3fffffff
typedef long long LL;
using namespace std;
int n;
bool tu[nn][nn];
int vis[nn];
bool dp[nn][nn][nn];
int pre[nn][nn][nn];
struct node
{
    int en,next;
}E[2*nn*nn];
int p[nn],num,cnt;
bool ok;
vector<int>ve[nn][2];
vector<int>ans[2];
void init()
{
    ans[0].clear();
    ans[1].clear();
    memset(p,-1,sizeof(p));
    memset(pre,-1,sizeof(p));
    memset(dp,false,sizeof(dp));
    num=0;
    cnt=0;
    ok=true;
    memset(vis,-1,sizeof(vis));
    for(int i=0;i<=n;i++)
    {
        ve[i][0].clear();
        ve[i][1].clear();
    }
}
void add(int st,int en)
{
    E[num].en=en;
    E[num].next=p[st];
    p[st]=num++;

    E[num].en=st;
    E[num].next=p[en];
    p[en]=num++;
}
void dfs(int id,int co)
{
    if(!ok)
        return ;
    vis[id]=co;
    ve[cnt][co].push_back(id);
    int i,w;
    for(i=p[id];i+1;i=E[i].next)
    {
        w=E[i].en;
        if(vis[w]==-1)
            dfs(w,1-co);
        else if(vis[w]==co)
        {
            ok=false;
        }
        if(!ok)
            return ;
    }
}
int main()
{
    int i,x,j,g;
    while(scanf("%d",&n)!=EOF)
    {
        memset(tu,false,sizeof(tu));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&x);
            while(x)
            {
                tu[i][x]=true;
                scanf("%d",&x);
            }
        }
        init();
        for(i=1;i<=n;i++)
        {
            for(j=i+1;j<=n;j++)
            {
                if(!tu[i][j]||!tu[j][i])
                {
                    add(i,j);
                }
            }
        }
        for(i=1;i<=n;i++)
        {
            if(vis[i]==-1)
            {
                cnt++;
                dfs(i,0);
                if(!ok)
                    break;
            }
        }
        if(!ok)
        {
            puts("No solution");
            continue;
        }
        dp[0][0][0]=true;
        int i1,i2;
        for(i=1;i<=cnt;i++)
        {
            i1=ve[i][0].size();
            i2=ve[i][1].size();
            for(j=0;j<=n;j++)
            {
                for(g=0;g+j<=n;g++)
                {
                    if(j-i1>=0&&g-i2>=0)
                    {
                        if(dp[i-1][j-i1][g-i2])
                        {
                            dp[i][j][g]=true;
                            pre[i][j][g]=0;
                            continue;
                        }
                    }
                    if(j-i2>=0&&g-i1>=0)
                    {
                        if(dp[i-1][j-i2][g-i1])
                        {
                            dp[i][j][g]=true;
                            pre[i][j][g]=1;
                        }
                    }
                }
            }
        }
        for(i=n/2;i>=0;i--)
        {
            if(dp[cnt][i][n-i])
            {
                break;
            }
        }
        i1=i,i2=n-i;
        while(cnt)
        {
            if(pre[cnt][i1][i2]==0)
            {
                i1-=ve[cnt][0].size();
                i2-=ve[cnt][1].size();
                for(i=0;i<(int)ve[cnt][0].size();i++)
                {
                    ans[0].push_back(ve[cnt][0][i]);
                }
                for(i=0;i<(int)ve[cnt][1].size();i++)
                {
                    ans[1].push_back(ve[cnt][1][i]);
                }
                cnt--;
            }
            else
            {
                i1-=ve[cnt][1].size();
                i2-=ve[cnt][0].size();
                for(i=0;i<(int)ve[cnt][1].size();i++)
                {
                    ans[0].push_back(ve[cnt][1][i]);
                }
                for(i=0;i<(int)ve[cnt][0].size();i++)
                {
                    ans[1].push_back(ve[cnt][0][i]);
                }
                cnt--;
            }
        }
        printf("%d",(int)ans[0].size());
        for(i=0;i<(int)ans[0].size();i++)
        {
            printf(" %d",ans[0][i]);
        }
        puts("");
        printf("%d",(int)ans[1].size());
        for(i=0;i<(int)ans[1].size();i++)
        {
            printf(" %d",ans[1][i]);
        }
        puts("");
    }
    return 0;
}