HDU 5876 补图的最短路

参考博客

题意：

给你一个图G，问补图上s到其他n-1个点的最短路。

题解：

bfs + set。每次把u点不相连的点入队列，取更新。具体见代码。

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, n) for(int i = a; i <= n; ++ i);
#define per(i, a, n) for(int i = n; i >= a; -- i);
typedef long long ll;
const int N = 2e5+ 5;
const int mod = 998244353;
const double Pi = acos(- 1.0);
const int INF = 0x3f3f3f3f;
const int G = 3, Gi = 332748118;
ll qpow(ll a, ll b) { ll res = 1; while(b){ if(b) res = (res * a) % mod; a = (a * a) % mod; b >>= 1;} return res; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b);}
bool cmp(int a, int b){ return a > b;}
//

int T, n, m, s;
int head[N], cnt = 0;
int dis[N];
queue<int> q;
set<int> ta, tb;


struct node{
    int to, nxt;
}edge[N << 1];

void add(int u, int v){
    edge[cnt].to = v, edge[cnt].nxt = head[u], head[u] = cnt ++;
    edge[cnt].to = u, edge[cnt].nxt = head[v], head[v] = cnt ++;
}

void bfs(){
    while(q.size()) q.pop();
    set<int>::iterator it;
    ta.clear(), tb.clear();
    dis[s] = 0;
    for(int i = 1; i <= n; ++ i) ta.insert(i);
    ta.erase(s);
    q.push(s);
    while(q.size()){
        int u = q.front(); q.pop();
        for(int i = head[u]; i != -1; i = edge[i].nxt){
            int v = edge[i].to;
            if(!ta.count(v)) continue;
            ta.erase(v);
            tb.insert(v);
        }
        for(it = ta.begin(); it != ta.end(); ++ it){
            q.push(*it);
            dis[*it] = dis[u] + 1;
        }
        ta = tb;
        tb.clear();
    }
}

int main()
{
    scanf("%d",&T);
    while(T --){
        cnt = 0;
        scanf("%d%d",&n,&m);
        for(int i = 0; i <= n; ++ i) head[i] = -1;
        for(int i = 1; i <= m; ++ i){
            int x, y; scanf("%d%d",&x,&y);
            add(x, y);
        }
        scanf("%d",&s);
        bfs();
        int num = 0;
        for(int i = 1; i <= n; ++ i){
            if(i == s) continue;
            num ++;
            if(dis[i] == INF) printf("-1");
            else printf("%d",dis[i]);
            if(num == n - 1) printf("\n");
            else printf(" ");
        }
    }
    return 0;
}