IndiaHacks 2016

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 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const int Mod = 1e9 + 7;

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[0].nx;
    Link[0].nx = tot++;
}

int main() {
    Link[0].nx = -1; tot = 1;
    scanf("%d%d%d", &n, &m, &k);
    cnt1 = 0;
    for(int i = 1; i <= m; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        ban[u].push_back(v);
        ban[v].push_back(u);
        if(u == 1) ok[v] = true;
        if(v == 1) ok[u] = true;
    }
    for(int i = 2; i <= n; i++) add(i);

    int down = 0, up = 0;
    while(Link[0].nx != -1) {
        int cnt = 0;
        queue<int> que;
        
        que.push(Link[Link[0].nx].id);
        Link[0].nx = Link[Link[0].nx].nx;

        while(!que.empty()) {
            int u = que.front(); que.pop();
            if(!ok[u]) cnt++;
            for(int b : ban[u]) vis[b] = true;
            pre = 0;
            for(int i = Link[0].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 0;
        }
        down++;
        up += cnt;
    }

    if(k >= down && k <= up) puts("possible");
    else puts("impossible");
    return 0;
}

/*
*/