UVaLive 4628 Jack's socks (贪心)

题意：给定一个无向图，让你把所有点的和它的任意一个相邻点匹配起来，问你是方案是不是唯一，如果是，则输出方案。

析：贪心，很容易知道，如果一个点的度数是 1，那么它只有一个相邻点，这样的话，我们就可以把它和它相邻结点匹配，然后把与它相邻结点也相邻的点的度数都减 1，然后再找度数为 1 的点，直接找不到为止，如果最后所得到的匹配数是 n / 2 个，那么方案就是唯一，否则不是无解，就是多种方案。

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>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#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 fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-10;
const int maxn = 1000 + 5;
const int maxm = 700 + 10;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -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;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }


vector<int> G[maxn];
int in[maxn];
vector<P> ans;
bool vis[maxn];

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; ++i)  G[i].cl;
    ms(in, 0);
    for(int i = 0; i < m; ++i){
      int a, b;  scanf("%d %d", &a, &b);
      G[a].pb(b);  G[b].pb(a); ++in[a];  ++in[b];
    }
    ans.cl;  ms(vis, 0);
    while(ans.sz != n / 2){
      bool ok = false;
      for(int i = 1; i <= n; ++i)  if(!vis[i] && in[i] == 1){
        ok = true;  vis[i] = 1;
        --in[i];
        for(int j = 0; j < G[i].sz; ++j){
          int v = G[i][j];
          if(!vis[v]){
            ans.pb(P(min(i, v), max(i, v)));
            vis[v] = 1;
            for(int k = 0; k < G[v].sz; ++k)  --in[G[v][k]];
            break;
          }
        }
      }
      if(!ok)  break;
    }
    if(ans.sz !=  n/2)  puts("NO");
    else{
      puts("YES");
      sort(ans.be, ans.ed);
      for(auto &it : ans)  printf("%d %d\n", it.fi, it.se);
    }
  }
  return 0;
}