【HDU5521 2015沈阳赛区M】【拆点最短路 dijkstra+heap】Meeting 集合内距离相同


Meeting

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 156    Accepted Submission(s): 45


Problem Description
Bessie and her friend Elsie decide to have a meeting. However, after Farmer John decorated his
fences they were separated into different blocks. John's farm are divided into   n  blocks labelled from   1  to   n.
Bessie lives in the first block while Elsie lives in the   n-th one. They have a map of the farm
which shows that it takes they   ti  minutes to travel from a block in   Ei  to another block
in   Ei  where   Ei (1im)  is a set of blocks. They want to know how soon they can meet each other
and which block should be chosen to have the meeting.
 

Input
The first line contains an integer   T (1T6), the number of test cases. Then   T  test cases
follow.

The first line of input contains   n  and   m.   2n105. The following   m  lines describe the sets   Ei (1im). Each line will contain two integers   ti(1ti109)  and   Si (Si>0)  firstly. Then   Si  integer follows which are the labels of blocks in   Ei. It is guaranteed that   mi=1Si106.
 

Output
For each test case, if they cannot have the meeting, then output "Evil John" (without quotes) in one line.

Otherwise, output two lines. The first line contains an integer, the time it takes for they to meet.
The second line contains the numbers of blocks where they meet. If there are multiple
optional blocks, output all of them in ascending order.
 

Sample Input
 
       
2 5 4 1 3 1 2 3 2 2 3 4 10 2 1 5 3 3 3 4 5 3 1 1 2 1 2
 

Sample Output
 
       
Case #1: 3 3 4 Case #2: Evil John
Hint
In the first case, it will take Bessie 1 minute travelling to the 3rd block, and it will take Elsie 3 minutes travelling to the 3rd block. It will take Bessie 3 minutes travelling to the 4th block, and it will take Elsie 3 minutes travelling to the 4th block. In the second case, it is impossible for them to meet.
 

Source
2015ACM/ICPC亚洲区沈阳站-重现赛(感谢东北大学)
 

 
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template  inline void gmax(T &a,T b){if(b>a)a=b;}
template  inline void gmin(T &a,T b){if(bb.v;}
};
void ins(int x,int y,int z)
{
	id++;
	w[id]=y;
	c[id]=z;
	nxt[id]=first[x];
	first[x]=id;
}
priority_queueq;
void inq(int x,LL dis)
{
	if(dis>=f[x])return;
	f[x]=dis;
	q.push(node(x,dis));
}
void dijkstra(int st)
{
	MS(f,63);
	MS(e,0);
	inq(st,0);
	q.push(node(st,0));
	while(!q.empty())
	{
		int x=q.top().x;q.pop();
		if(e[x])continue;e[x]=1;
		for(int z=first[x];z;z=nxt[z])inq(w[z],f[x]+c[z]);
	}
}
int main()
{
	scanf("%d",&casenum);
	for(casei=1;casei<=casenum;casei++)
	{
		MS(first,0);id=1;
		int n,m;
		scanf("%d%d",&n,&m);
		for(int i=1;i<=m;i++)
		{
			int dis,g,x;
			scanf("%d%d",&dis,&g);
			int in=n+i;
			int out=n+m+i;
			ins(in,out,dis);
			while(g--)
			{
				scanf("%d",&x);
				ins(x,in,0);
				ins(out,x,0);
			}
		}
		dijkstra(1);
		MC(F,f);
		dijkstra(n);
		LL ans=1e18;
		int ed;
		for(int i=1;i<=n;i++)
		{
			gmax(F[i],f[i]);
			if(F[i]<=ans)
			{
				ans=F[i];
				ed=i;
			}
		}
		if(ans==1e18)printf("Case #%d: Evil John\n",casei);
		else
		{
			printf("Case #%d: %lld\n",casei,ans);
			for(int i=1;i


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