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POJ 1112 Team Them Up! 二分图判定+01背包

2016-10-01 12:42 369 查看

题目链接：

Team Them Up!

Time Limit: 1000MSMemory Limit: 10000K

问题描述

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.

输入

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.

输出

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.

样例输入

5

2 3 5 0

1 4 5 3 0

1 2 5 0

1 2 3 0

4 3 2 1 0

样例输出

3 1 3 5

2 2 4

题意

给你若干个人，要把他们分成两组，其中同一组内要保证任意两个人互相认识（题目给的边都是单向边），求一个使得两组人数最接近的方案，要求输出一个可行分组方案。

题解

首先，把关系图建处理（对于(u,v)如果不存在(v,u)相当于这条边不存在），处理出补图，然后题目就转换成了一个二分图问题了，跑下黑白染色，判断是否有可行解，如果有，则用dp跑下，求出最优解就ok啦。dp[i][j]表示已经处理了i个联通分量（跑黑白染色会处理出若干个联通分量），其中一组能凑出j个人的方案。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<sstream>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=111;

int n;
bool gra[maxn][maxn];
bool dp[maxn][maxn];
VI lis[maxn][3];
int color[maxn],tot;
//黑白染色
bool dfs(int u) {
for(int v=1; v<=n; v++) {
if(!gra[u][v]) continue;
if(!color[v]) {
color[v]=3-color[u];
lis[tot][color[v]].pb(v);
bool su=dfs(v);
if(!su) return false;
} else {
if(color[v]==color[u]) return false;
}
}
return true;
}

VI ans[3];
bool used[maxn];
//处理出最优方案
void print(int i,int j) {
if(i==0) {
return;
}
int a=lis[i][1].sz(),b=lis[i][2].sz();
if(j>=a&&dp[i-1][j-a]) {
print(i-1,j-a);
rep(ii,0,lis[i][1].sz()) {
ans[1].pb(lis[i][1][ii]);
}
} else if(j>=b&&dp[i-1][j-b]) {
print(i-1,j-b);
rep(ii,0,lis[i][2].sz()) {
ans[1].pb(lis[i][2][ii]);
}
}
}

int main() {
while(scf("%d",&n)==1&&n) {
clr(gra,0);
for(int i=1; i<=n; i++) {
int v;
while(scf("%d",&v)==1&&v) {
gra[i][v]=1;
}
}
for(int i=1; i<=n; i++) {
for(int j=1; j<i; j++) {
gra[i][j]=gra[j][i]=!(gra[i][j]&&gra[j][i]);
}
}

//黑白染色
clr(color,0);
rep(i,0,maxn) lis[i][1].clear(),lis[i][2].clear();
tot=0;
bool su=1;
for(int i=1; i<=n; i++) {
if(!color[i]) {
++tot;
color[i]=1;
lis[tot][1].pb(i);
su=dfs(i);
}
if(!su) break;
}
if(!su) {
prf("No solution\n");
continue;
}

//01背包，不是选和不选，而是选1还是选2.
clr(dp,0);
dp[0][0]=1;
for(int i=1; i<=tot; i++) {
int a=lis[i][1].sz(),b=lis[i][2].sz();
for(int j=0; j<=100; j++) {
if(j>=a&&j>=b) {
dp[i][j]=dp[i-1][j-a]|dp[i-1][j-b];
} else if(j>=a) {
dp[i][j]=dp[i-1][j-a];
} else if(j>=b) {
dp[i][j]=dp[i-1][j-b];
}
}
}

//输出最优方案
int Mi=INF,pos=-1;
for(int i=1; i<=100; i++) {
if(dp[tot][i]) {
if(Mi>abs(i-(n-i))) {
Mi=abs(i-(n-i));
pos=i;
}
}
}

ans[1].clear(),ans[2].clear();
clr(used,0);
print(tot,pos);

rep(i,0,ans[1].sz()) used[ans[1][i]]=1;
for(int i=1; i<=n; i++) {
if(!used[i]) ans[2].pb(i);
}

sort(all(ans[1]));
sort(all(ans[2]));

for(int i=1; i<=2; i++) {
prf("%d ",ans[i].sz());
rep(j,0,ans[i].sz()) prf("%d%c",ans[i][j],j==ans[i].sz()-1?'\n':' ');
}
}
return 0;
}

//end-----------------------------------------------------------------------

/*
3
2 0
1 3 0
2 0
*/

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