UVa 1627 Team them up! (01背包+二分图)

题意：给n个分成两个组，保证每个组的人都相互认识，并且两组人数相差最少，给出一种方案。

析：首先我们可以知道如果某两个人不认识，那么他们肯定在不同的分组中，所以我们可以根据这个结论构造成一个图，如果两个不相互认识，

那么就加一条边，然后如果这个图是二分图，那么这分组是可以，否则就是不可能的。然后dp[i][j]表示那两个组相差人数为 j 是不是可以达到，

当然可能为负数，所以可以提前加上n，然后就是逆序输出答案即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<double, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 100 + 10;
const int mod = 1e6;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int color[maxn];
bool g[maxn][maxn];
vector<int> v[maxn][2];
int diff[maxn], cnt;
bool dp[maxn][maxn*2];

bool dfs(int u, int val){
  color[u] = val;
  v[cnt][val-1].push_back(u);
  for(int i = 1; i <= n; ++i){
    if(i == u)  continue;
    if(g[u][i] && g[i][u])   continue;
    if(color[i] == color[u])  return false;
    if(!color[i] && !dfs(i, 3-val))  return false;
  }
  return true;
}

void print(int ans){
  vector<int> v1, v2;
  for(int i = cnt-1; i >= 0; --i){
    int t;
    if(dp[i][ans+n+diff[i]]){ t = 0; ans += diff[i]; }
    else { t = 1;  ans -= diff[i]; }
    for(int j = 0; j < v[i][t].size(); ++j)
      v1.push_back(v[i][t][j]);
    for(int j = 0; j < v[i][t^1].size(); ++j)
      v2.push_back(v[i][t^1][j]);
  }
  printf("%d", v1.size());
  for(int i = 0; i < v1.size(); ++i)  printf(" %d", v1[i]);
  printf("
");
  printf("%d", v2.size());
  for(int i = 0; i < v2.size(); ++i)  printf(" %d", v2[i]);
  printf("
");
}

void solve(){
  memset(dp, 0, sizeof dp);
  dp[0][n] = true;
  for(int i = 0; i < cnt; ++i){
    for(int j = -n; j <= n; ++j)  if(dp[i][j+n]){
      dp[i+1][j+n+diff[i]] = true;
      dp[i+1][j+n-diff[i]] = true;
    }
  }
  for(int i = 0; i <= n; ++i){
    if(dp[cnt][i+n]){ print(i);  return ; }
    if(dp[cnt][-i+n]){ print(-i);  return ; }
  }
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    memset(g, 0, sizeof g);
    for(int i = 1; i <= n; ++i){
      while(scanf("%d", &m) == 1 && m) g[i][m] = true;
    }
    memset(color, 0, sizeof color);
    bool ok = true;
    cnt = 0;
    for(int i = 1; i <= n && ok; ++i)  if(!color[i]){
      v[cnt][0].clear();
      v[cnt][1].clear();
      ok = dfs(i, 1);
      diff[cnt] = (int)v[cnt][0].size() - v[cnt][1].size();
      ++cnt;
    }
    if(!ok)  puts("No solution");
    else solve();
    if(T)  printf("
");
  }
  return 0;
}