HDOJ5521(巧妙构建完全图)

Posted 2020-08-16 vCoders

Meeting

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2415    Accepted Submission(s): 765

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.

 

 

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.

 

 

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

 

思路:任意取一个block举例建图。设block中有n个节点,，有题意可知，block是一个无向完全图。若按传统的方法建图，则无向完全图中有n个节点, n*(n-1)条边。我们换一种方式建图，在block中增加两个节点u，v。再加入一条有向边（u，v），然后将block中的所有节点xi，加入两条有向边（xi，u）、（v，xi）。最终得到的是一幅有n+2个节点2*n+1个有向边的有向图，其等价于按传统方法构建的完全图。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXN = 400005;
const int INF = 0x3f3f3f3f;
struct Edge{
    int to, w, net;
}es[4000005];
int head[MAXN], tot;
int n, m;
void addedge(int u, int v, int w)
{
    es[tot].to = v;
    es[tot].w = w;
    es[tot].net = head[u];
    head[u] = tot++;
}
void spfa(int src, int d[], bool vis[], int n)
{
    for(int i = 1; i <= n; i++)
    {
        d[i] = INF;
        vis[i] =false;
    }
    d[src] = 0;
    queue<int> que;
    que.push(src);
    while(!que.empty())
    {
        int u = que.front(); que.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1; i = es[i].net)
        {
            Edge e = es[i];
            if(d[e.to] > d[u] + e.w)
            {
                d[e.to] = d[u] + e.w;
                if(!vis[e.to])
                {
                    vis[e.to] = true;
                    que.push(e.to);
                }
            }
        }
    }
}
int d[2][MAXN];
bool vis[MAXN];
int vec[MAXN], len;
int main()
{
    int T;
    scanf("%d", &T);
    for(int cas = 1; cas <= T; cas++)
    {
        memset(head, -1, sizeof(head));
        tot = 0;
        scanf("%d %d", &n ,&m);
        int newN = n;
        for(int i = 0; i < m; i++)
        {
            int w, s;
            scanf("%d %d", &w, &s);
            //巧妙构图
            int u = ++newN;
            int v = ++newN;
            addedge(u, v, w);
            for(int j = 0; j < s; j++)
            {
                int x;
                scanf("%d", &x);
                addedge(x, u, 0);
                addedge(v, x, 0);
            }
        }

        spfa(1, d[0], vis, newN);
        spfa(n, d[1], vis, newN);

        int mn = INF;
        for(int i = 1; i <= n; i++)
        {
            int mx = max(d[0][i], d[1][i]);
            if(mn > mx)
            {
                mn = mx;
            }
        }
        if(mn == INF)
        {
            printf("Case #%d: Evil John\n", cas);
            continue;
        }
        len = 0;
        for(int i = 1; i <= n; i++)
        {
            if(max(d[0][i], d[1][i]) == mn)
            {
                vec[len++] = i;
            }
        }
        printf("Case #%d: %d\n", cas, mn);
        for(int i = 0; i < len -1; i++)
        {
            printf("%d ", vec[i]);
        }
        printf("%d\n", vec[len-1]);
    }
    return 0;
}