HDU5521 Meeting【最短路】

题意：一张图，双向边，A在1节点，B在n节点，两人都能动，问他们最短的相遇时间，可以哪些点碰面。给了很多集合，集合内的点相互到达的时间为t

思路：给每个集合建一个点，集合内的点都连到这个点上，分别从A、B跑spfa，取每个点的两次最短距离的最大值，就是在这个点最短的遇见时间。巧妙建图，想不出QWQ

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<math.h>
#include<vector>
#include<list>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<algorithm>
#include<numeric>
#include<functional>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 1e6+5;
ll d[maxn],d2[maxn];
int t[maxn],inq[maxn];
int ans[maxn],go;

struct Edge
{
	int to,next;
	ll val;
}edge[maxn*2];
int head[maxn],tot;

void add(int x,int y,ll val)
{
	edge[tot].to = y;
	edge[tot].val = val;
	edge[tot].next = head[x];
	head[x] = tot++;
}

void init(int x)
{
	memset(head,-1,sizeof head);
	tot = 0;
}

void spfa1(int x)
{
	queue<int> q;
	while(!q.empty())
		q.pop();
	memset(inq,0,sizeof inq);
	memset(d,0x3f,sizeof d);
	d[x] = 0;
	q.push(x);
	while(!q.empty())
	{
		int now = q.front();q.pop();
		inq[now] = 0;
		for(int i = head[now]; i != -1; i = edge[i].next)
		{
			int temp = edge[i].to;
			if(d[now] + edge[i].val < d[temp])
			{
				d[temp] = d[now] + edge[i].val;
				if(!inq[temp])
				{
					inq[temp] = 1;
					q.push(temp);
				}
			}
		}
	}
}

void spfa2(int x)
{
	queue<int> q;
	while(!q.empty())
		q.pop();
	memset(inq,0,sizeof inq);
	memset(d2,0x3f,sizeof d2);
	d2[x] = 0;
	q.push(x);
	while(!q.empty())
	{
		int now = q.front();q.pop();
		inq[now] = 0;
		for(int i = head[now]; i != -1; i = edge[i].next)
		{
			int temp = edge[i].to;
			if(d2[now] + edge[i].val < d2[temp])
			{
				d2[temp] = d2[now] + edge[i].val;
				if(!inq[temp])
				{
					inq[temp] = 1;
					q.push(temp);
				}
			}
		}
	}
}

int main(void)
{
	int n,m,T,kase = 1;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		init(n);
		for(int i = 1; i <= m; i++)
		{
			int tt,tp;
			ll val;
			scanf("%lld%d",&val,&tt);
			for(int j = 1; j <= tt; j++)
			{
				scanf("%d",&tp);
				add(tp,i+n,val);
				add(i+n,tp,val);
			}
		}
		spfa1(1);
		spfa2(n);
		ll time = 0x3f3f3f3f3f3f3f3f;
		go = 0;
		for(int i = 1; i <= n; i++)
		{
			if(max(d[i],d2[i]) < time)
			{
				time = max(d[i],d2[i]);
				go = 0;
				ans[go++] = i;
			}
			else if(max(d[i],d2[i]) == time && time != 0x3f3f3f3f3f3f3f3f)
			{
				ans[go++] = i;
			}
		}
		if(go == 0)
		{
			printf("Case #%d: Evil John\n",kase++);
			continue;
		}
		printf("Case #%d: %lld\n",kase++,time/2);
		for(int i = 0; i < go; i++)
		{
			printf("%d%c",ans[i],i == go-1?'\n':' ');
		}
	}
	return 0;
}