HDU5521 Meeting(最短路建圖，堆優化的dijkstra)

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

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.

思路

題目把$n$個點分成了$m$個塊，規定每一個塊裏面的任意兩點相互到達的時間都爲$t_i$,現在有兩個人，一個從$1$點出發，另一個從$n$點出發，他們要找一個點開會，這個點要滿足花費的時間最小。題目問你滿足時間花費最小的時間是多少，並且把滿足這些時間的所有點輸出。

這個題如果給每個塊的任意兩點都建邊的話肯定無用的邊會很多，這個題需要一點建圖思路，我們對於每一個塊，建立一個虛點，讓塊中每個點都連上自己這個塊中建立的虛點，權值就爲這個塊的$t_i$，那我從起點跑一遍最短路，從終點跑一遍最短路，只需要找到這些點中最小的時間花費的這個點即可ans = min(ans, max(dis1[i], dis2[i])).

配合快讀食用更加

代碼

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mem(a, b) memset(a, b, sizeof(a))
namespace IO //使用的時候先IO::begin()
{            //然後輸入時IO::read(x)
const ll MX = 4e7;
char buf[MX];
ll c, sz;
void begin()
{
    c = 0;
    sz = fread(buf, 1, MX, stdin);
}
inline bool read(ll &t)
{
    while (c < sz && buf[c] != '-' && (buf[c] < '0' || buf[c] > '9'))
        c++;
    if (c >= sz)
        return false;
    bool flag = 0;
    if (buf[c] == '-')
        flag = 1, c++;
    for (t = 0; c < sz && '0' <= buf[c] && buf[c] <= '9'; c++)
        t = t * 10 + buf[c] - '0';
    if (flag)
        t = -t;
    return true;
}
} // namespace IO
const ll N = 1e6 + 10;
const ll inf = 1e18 + 10;
ll n, m;
ll first[N], tot;
ll dis1[N], vis[N], dis2[N];
struct edge
{
    ll v, w, next;
} e[N * 2];
struct node
{
    ll id, now;
    node() {}
    node(ll _id, ll _now)
    {
        id = _id;
        now = _now;
    }
    bool friend operator<(node a, node b)
    {
        return a.now > b.now;
    }
};
void dijkstra1(ll st)
{
    for (ll i = 1; i <= n + m; i++)
    {
        dis1[i] = inf;
        vis[i] = 0;
    }
    dis1[st] = 0;
    priority_queue<node> q;
    q.push(node(st, 0));
    while (!q.empty())
    {
        node u = q.top();
        q.pop();
        if (!vis[u.id])
        {
            vis[u.id] = 1;
            for (ll i = first[u.id]; ~i; i = e[i].next)
            {
                ll v = e[i].v, w = e[i].w;
                if (dis1[u.id] + w < dis1[v])
                {
                    dis1[v] = dis1[u.id] + w;
                    q.push(node(v, dis1[v]));
                }
            }
        }
    }
}
void dijkstra2(ll st)
{
    for (ll i = 1; i <= n + m; i++)
    {
        dis2[i] = inf;
        vis[i] = 0;
    }
    dis2[st] = 0;
    priority_queue<node> q;
    q.push(node(st, 0));
    while (!q.empty())
    {
        node u = q.top();
        q.pop();
        if (!vis[u.id])
        {
            vis[u.id] = 1;
            for (ll i = first[u.id]; ~i; i = e[i].next)
            {
                ll v = e[i].v, w = e[i].w;
                if (dis2[u.id] + w < dis2[v])
                {
                    dis2[v] = dis2[u.id] + w;
                    q.push(node(v, dis2[v]));
                }
            }
        }
    }
}

void add_edge(ll u, ll v, ll w)
{
    e[tot].v = v, e[tot].w = w;
    e[tot].next = first[u];
    first[u] = tot++;
}
void init()
{
    mem(first, -1);
    tot = 0;
}
void solve()
{
    ll u, w, x, t;
    IO::read(n), IO::read(m);
    init();
    for (ll i = 1; i <= m; i++)
    {
        IO::read(w), IO::read(t);
        while (t--)
        {
            IO::read(u);
            add_edge(u, n + i, w);
            add_edge(n + i, u, w);
        }
    }
    dijkstra1(1);
    dijkstra2(n);
    ll ans = inf;
    for (ll i = 1; i <= n; i++)
    {
        dis1[i] /= 2;
        dis2[i] /= 2;
        ans = min(ans, max(dis1[i], dis2[i]));
    }
    if (ans * 2 == inf)
        puts("Evil John");
    else
    {
        printf("%lld\n", ans);
        ll flag = 1;
        for (ll i = 1; i <= n; i++)
        {
            if (ans == max(dis1[i], dis2[i]))
            {
                if (flag)
                {
                    printf("%lld", i);
                    flag = 0;
                }
                else
                    printf(" %lld", i);
            }
        }
        puts("");
    }
}
int main()
{
    // freopen("in.txt", "r", stdin);
    IO::begin();
    ll t, q = 1;
    IO::read(t);
    while (t--)
    {
        printf("Case #%lld: ", q++);
        solve();
    }
    return 0;
}