E - Bear and Forgotten Tree 2

思路:先不考虑1这个点,求有多少个连通块,每个连通块里有多少个点能和1连,这样就能确定1的度数的上下界。

求连通块用链表维护。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std; const int N = 3e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = ;
const int Mod = 1e9 + ; int n, m, k, tot, pre, cnt1, cnt2;
bool vis[N], ok[N];
vector<int> ban[N]; struct node {
int id, nx;
} Link[N]; void add(int id) {
Link[tot].id = id;
Link[tot].nx = Link[].nx;
Link[].nx = tot++;
} int main() {
Link[].nx = -; tot = ;
scanf("%d%d%d", &n, &m, &k);
cnt1 = ;
for(int i = ; i <= m; i++) {
int u, v;
scanf("%d%d", &u, &v);
ban[u].push_back(v);
ban[v].push_back(u);
if(u == ) ok[v] = true;
if(v == ) ok[u] = true;
}
for(int i = ; i <= n; i++) add(i); int down = , up = ;
while(Link[].nx != -) {
int cnt = ;
queue<int> que; que.push(Link[Link[].nx].id);
Link[].nx = Link[Link[].nx].nx; while(!que.empty()) {
int u = que.front(); que.pop();
if(!ok[u]) cnt++;
for(int b : ban[u]) vis[b] = true;
pre = ;
for(int i = Link[].nx; ~i; i = Link[i].nx) {
int to = Link[i].id;
if(!vis[to]) {
que.push(to);
Link[pre].nx = Link[i].nx;
} else pre = i;
}
for(int b : ban[u]) vis[b] = false;
}
if(!cnt) {
puts("impossible");
return ;
}
down++;
up += cnt;
} if(k >= down && k <= up) puts("possible");
else puts("impossible");
return ;
} /*
*/

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