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("\n");

  printf("%d", v2.size());

  for(int i = 0; i < v2.size(); ++i)  printf(" %d", v2[i]);

  printf("\n");

}

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("\n");

  }

  return 0;

}

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UVa 1627 Team them up! (01背包+二分图)

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