Meeting 加虚拟边

最新推荐文章于 2025-03-31 12:22:20 发布

转载最新推荐文章于 2025-03-31 12:22:20 发布 · 120 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/joeylee97/p/7395535.html

在Farmer John将农场分为多个区块后，Bessie和Elsie试图找到最快的方式在某一点会面。通过构建一个包含各区块间旅行时间的图，并使用SPFA算法寻找最短路径，此问题得以解决。

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 nn blocks labelled from 11 to nn.
Bessie lives in the first block while Elsie lives in the nn-th one. They have a map of the farm
which shows that it takes they titi minutes to travel from a block in EiEito another block
in EiEi where Ei (1≤i≤m)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.

InputThe first line contains an integer T (1≤T≤6)OutputFor 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

Sample Output

Case #1: 3
3 4
Case #2: Evil John

可以将给顶集合的元素连到一个虚拟结点上，求出最短路来再/2,这样避免了大量的重复加边，还避免了小数

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<memory>
#include<bitset>
#include<string>
#include<functional>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MAXN = 5e5 ;

#define INF 0x3f3f3f3f

/*
连接虚拟结点
到该点的距离为L
求出最短路/2 避免小数！
*/
LL T, d, n, m, cnt;
struct edge
{
    edge(LL _a,LL _b):to(_a),cost(_b){}
    LL to, cost;
};
vector<edge>E[MAXN];
LL    dist1[MAXN], dist2[MAXN];
bool vis[MAXN];
void addedge(LL f,LL to,LL dis)
{
    E[f].push_back(edge(to, dis));
    E[to].push_back(edge(f, dis));
}
void init()
{
    for (LL i = 0; i < MAXN; i++)
        E[i].clear();
}
void spfa(LL beg, LL lowcost[])
{
    queue<LL> q;
    memset(vis, false, sizeof(vis));
    for (int i = 0; i <= n + m; i++)
        lowcost[i] = INF;
    lowcost[beg] = 0;
    vis[beg] = true;
    q.push(beg);
    while (!q.empty())
    {
        LL f = q.front();
        q.pop();
        vis[f] = false;
        for (int i = 0; i < E[f].size(); i++)
        {
            if (lowcost[E[f][i].to] > lowcost[f] + E[f][i].cost)
            {
                lowcost[E[f][i].to] = lowcost[f] + E[f][i].cost;
                if (!vis[E[f][i].to])
                {
                    vis[E[f][i].to] = true;
                    q.push(E[f][i].to);
                }
            }
        }
    }
}
int main()
{
    scanf("%lld", &T);
    for(LL cas = 1;cas <= T; cas++)
    {
        init();
        scanf("%lld%lld", &n, &m);
        LL tmp, tt;
        for (LL i = 1; i <= m; i++)
        {
            scanf("%lld%lld", &d, &tmp);
            while (tmp--)
            {
                scanf("%lld", &tt);
                addedge(tt, n + i, d);
            }
        }
        spfa(1, dist1);
        spfa(n , dist2);
        LL ans = INF;
        for (int i = 0; i <= n; i++)
            ans = min(ans, max(dist1[i], dist2[i]));
        if (ans == INF)
            printf("Case #%lld: Evil John\n", cas);
        else
        {
            printf("Case #%lld: %lld\n", cas, ans / 2);
            bool f = false;
            for (int i = 1; i <= n; i++)
            {
                if (max(dist1[i], dist2[i]) == ans)
                {
                    if (!f)
                        printf("%d", i), f = true;
                    else
                        printf(" %d", i);
                }
            }
            printf("\n");
        }
    }
}

转载于:https://www.cnblogs.com/joeylee97/p/7395535.html