分析:构造题。可以这么想:先把s,t两个点去掉,把剩下的点先并查集合并。这样会出现
代码:
/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define offcin ios::sync_with_stdio(false)
#define sigma_size 26
#define lson l,m,v<<1
#define rson m+1,r,v<<1|1
#define slch v<<1
#define srch v<<1|1
#define sgetmid int m = (l+r)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define pb push_back
#define fi first
#define se second
const int INF = 0x3f3f3f3f;
const LL INFF = 1e18;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const int maxmat = 10;
const ull BASE = 31;
/*****************************************************/
typedef pair<int, int> pii;
const int maxn = 2e5 + 5;
int par[maxn];
int s, t, ds, dt;
bool inq[maxn];
int cons[maxn], cont[maxn];
std::vector<pii> vis;
std::vector<int> conn;
std::vector<int> G[maxn];
int N, M;
void init() {
for (int i = 0; i <= N; i ++)
par[i] = i;
}
int findpar(int x) {
return par[x] = (par[x] == x ? x : findpar(par[x]));
}
void unite(int x, int y) {
int px = findpar(x), py = findpar(y);
if (px == py) return;
// cout<<x<<" "<<y<<endl;
vis.pb(mk(x, y));
par[px] = py;
}
bool work() {
init();
for (int u = 1; u <= N; u ++) {
if (u == s || u == t) continue;
for (unsigned i = 0; i < G[u].size(); i ++) {
int v = G[u][i];
if (v == s || v == t) continue;
unite(u, v);
}
}
mem(inq, false);
for (int u = 1; u <= N; u ++) {
if (u == s || u == t) continue;
int p = findpar(u);
if (!inq[p]) {
conn.pb(p);
inq[p] = true;
}
}
mem(cons, -1); mem(cont, -1);
bool st = false;
for (unsigned i = 0; i < G[s].size(); i ++) {
int v = G[s][i];
if (v == t) {st = true; continue;}
int p = findpar(v);
cons[p] = v;
}
for (unsigned i = 0; i < G[t].size(); i ++) {
int v = G[t][i];
if (v == s) {st = true; continue;}
int p = findpar(v);
cont[p] = v;
}
int anss = 0, anst = 0;
for (unsigned i = 0; i < conn.size(); i ++) {
int p = conn[i];
if (cons[p] != -1 && cont[p] != -1) continue;
if (cons[p] == -1 && cont[p] == -1) return false;
if (cons[p] != -1 && cont[p] == -1) {
unite(s, cons[p]);
anss ++;
}
if (cons[p] == -1 && cont[p] != -1) {
unite(t, cont[p]);
anst ++;
}
}
if (anss >= ds || anst >= dt) return false;
ds -= anss, dt -= anst;
bool flag = true;
for (unsigned i = 0; i < conn.size(); i ++) {
int p = conn[i];
if (!(cons[p] != -1 && cont[p] != -1)) continue;
if (flag) {
ds --, dt --;
unite(s, cons[p]);
unite(t, cont[p]);
flag = false;
}
else {
if (ds) {
ds --;
unite(s, cons[p]);
}
else if (dt) {
dt --;
unite(t, cont[p]);
}
else return false;
}
}
int ps = findpar(s), pt = findpar(t);
if (ps != pt) {
if (ds && dt && st) unite(s, t);
else return false;
}
return true;
}
int main(int argc, char const *argv[]) {
cin>>N>>M;
for (int i = 0; i < M; i ++) {
int u, v; cin>>u>>v;
G[u].pb(v); G[v].pb(u);
}
cin>>s>>t>>ds>>dt;
if (work()) {
puts("Yes");
for (unsigned i = 0; i < vis.size(); i ++)
printf("%d %d\n", vis[i].fi, vis[i].se);
}
else puts("No");
return 0;
}