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  • Meeting 加虚拟边

    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 (1im)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 (1T6)T (1≤T≤6), the number of test cases. Then TT test cases 
    follow. 

    The first line of input contains nn and mm. 2n1052≤n≤105. The following mm lines describe the sets Ei (1im)Ei (1≤i≤m). Each line will contain two integers ti(1ti109)ti(1≤ti≤109) and Si (Si>0)Si (Si>0) firstly. Then SiSi integer follows which are the labels of blocks in EiEi. It is guaranteed that mi=1Si106∑i=1mSi≤106.
    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

    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
    

    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
    ", cas);
            else
            {
                printf("Case #%lld: %lld
    ", 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("
    ");
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/joeylee97/p/7395535.html
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