Meeting
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1358 Accepted Submission(s): 435
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 (1≤i≤m) 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.
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 (1≤i≤m) 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 (1≤T≤6),
the number of test cases. Then T test
cases
follow.
The first line of input contains n and m. 2≤n≤105. The following m lines describe the sets Ei (1≤i≤m). Each line will contain two integers ti(1≤ti≤109)and Si (Si>0) firstly. Then Si integer follows which are the labels of blocks in Ei. It is guaranteed that ∑mi=1Si≤106.
follow.
The first line of input contains n and m. 2≤n≤105. The following m lines describe the sets Ei (1≤i≤m). Each line will contain two integers ti(1≤ti≤109)and Si (Si>0) firstly. Then Si integer follows which are the labels of blocks in Ei. It is guaranteed that ∑mi=1Si≤106.
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.
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
题目链接：HDU 5521
题意：给定点数n和集合个数m，然后给你m个集合，每一个集合有si个点，两两之间的到达时间都是ti，一个人在1，一个人在n，求两人同时出发的相遇的最短时间
由于每一个集合的点有很多，若集合两两之间连边，边数非常大，一开始这样就超时了……然后正确做法是对每一个集合再虚拟一个节点（范围是[n+1,n+m]），给每一个集合内的点连边权为ti的双向边到本集合对应虚拟节点，集合内其他点不连边，这样就可以通过虚拟的节点来到达其他地方从而减少边数（方法真是太巧妙了），然后求相遇的最短时间，显然现在无法得知到底选哪个地点作为见面地点，那就对1跑一遍单源最短路，对n跑一边单源最短路，然后统计1~n中每一个点的可能最短时间（一个人早到一个人晚到，显然用max取时间长的那个数），然后选出1~n中的最短时间mndx，再遍历一下看哪些点的最短时间为mndx并记录输出，最后记得把最短时间除以2，因为连的边是ti，进入虚拟节点又出来会多算一次
点数为1e5，题目中说了所有集合大小之和不会超过1e6，每一个集合都有2*|Si|条边，那就是2*1e6条边因此N可设为1e5+10，M可设为2*1e6+10。
以上是以前的解法，昨天计蒜客被惨虐之后仔细看了一下D题发现其实跟这道题是同一个原理，这题是群内的点之间两两距离为ti，那不妨把Block点拆成入口和出口，然后这样连边：
<u, Block入口, 0>,<Block出口, u , 0>,<Block入口, Block出口, ti>
然后从S和T各跑一遍SPFA然后记录下Max更新答案即可，也不用像上面的解法一样除以2了，点数最差情况下一个点作为Block，应该是3e5，边数应该为2sum{Si}+m，应该是2e6+1e5
代码：
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 3e5 + 7;
const int M = 2e6 + 1e5 + 7;
struct edge
{
int to, nxt;
LL w;
edge() {}
edge(int _to, int _nxt, LL _w): to(_to), nxt(_nxt), w(_w) {}
};
edge E[M];
int head[N], tot;
int vis[N];
LL ds[N], de[N], Mindist[N];
void init()
{
CLR(head, -1);
tot = 0;
}
inline void add(int s, int t, LL d)
{
E[tot] = edge(t, head[s], d);
head[s] = tot++;
}
void spfa(int s, int flag, LL d[])
{
CLR(vis, 0);
if (flag)
CLR(ds, INF), ds[s] = 0;
else
CLR(de, INF), de[s] = 0;
queue<int>Q;
Q.push(s);
while (!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = 0;
for (int i = head[u]; ~i; i = E[i].nxt)
{
int v = E[i].to;
if (d[v] > d[u] + E[i].w)
{
d[v] = d[u] + E[i].w;
if (!vis[v])
{
vis[v] = 1;
Q.push(v);
}
}
}
}
}
int main(void)
{
int tcase, n, m, si, u, i;
scanf("%d", &tcase);
for (int q = 1; q <= tcase; ++q)
{
init();
scanf("%d%d", &n, &m);
LL ti;
for (i = 1; i <= m; ++i)
{
scanf("%I64d%d", &ti, &si);
add(i, i + m, ti); //m
while (si--)
{
scanf("%d", &u);
add(u + (m << 1), i, 0LL); //si
add(i + m, u + (m << 1), 0LL); //si
}
}
spfa(1 + (m << 1), 1, ds);
spfa(n + (m << 1), 0, de);
LL ans = 0x3f3f3f3f3f3f3f3f;
printf("Case #%d: ", q);
vector<int>pos;
for (i = 2 * m + 1; i <= 2 * m + n; ++i)
{
Mindist[i] = max<LL>(ds[i], de[i]);
if (Mindist[i] < ans)
ans = Mindist[i];
}
if (ans == 0x3f3f3f3f3f3f3f3f)
puts("Evil John");
else
{
printf("%I64d
", ans);
for (i = 2 * m + 1; i <= 2 * m + n; ++i)
if (Mindist[i] == ans)
pos.push_back(i - (m << 1));
int sz = pos.size();
for (i = 0; i < sz; ++i)
printf("%d%c", pos[i], "
"[i == sz - 1]);
}
}
return 0;
}