分糖果

本题的边权都为1，且小朋友吃糖果和传递糖果是并行的，所以答案为$距离发糖果小朋友最远的孩子拿到糖果的时间 + 吃糖果的时间$,故可直接采用bfs破本题

#include <iostream>
#include <cstring>
#include <queue>

using namespace std;

const int N = 2e6 + 10;

int n, p, c;
int m;
int h[N],e[N],ne[N],idx;
int dist[N];
bool v[N];

void add(int a,int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

int bfs()
{
    queue<int> q;
    q.push(c);
    dist[c] = 1;
    v[c] = true;

    while(q.size())
    {
        int t = q.front();
        q.pop();

        for(int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if(!v[j])
            {
                dist[j] = dist[t] + 1;
                q.push(j);
                v[j] = true;
            }
        }
    }

    int res = 0;
    for(int i = 1; i <= n; i ++ )
        res = max(res,dist[i]);
    return res + m;
}

int main()
{
    memset(h, -1, sizeof h);
    scanf("%d%d%d",&n,&p,&c);
    scanf("%d",&m);
    for(int i = 0; i < p; i ++ )
    {
        int a,b;
        scanf("%d%d",&a,&b);
        add(a,b);
        add(b,a);
    }

    int t = bfs();

    printf("%d",t);

    return 0;
}