• [Codeforces 666B] World Tour


    [题目链接]

             https://codeforces.com/contest/666/problem/B

    [算法]

           首先 , 用BFS求出任意两点的最短路径

           然后 , 我们用f[i][0-2]表示从i出发到达的最远三点 , g[i][0-2]表示到i距离最远的三个点

           枚举b和c , 然后在枚举3 * 3个点对 , 从中选出最优的a和d即可

           时间复杂度 : O(N^2)

    [代码]

             

    #include<bits/stdc++.h>
    using namespace std;
    #define MAXN 3010
    #define MAXM 5010
    const int inf = 2e9;
    
    int tot , n , m;
    int head[MAXN];
    int f[MAXN][3],g[MAXN][3],dist[MAXN][MAXN];
    
    struct info
    {
            int a , b , c , d;
    } res;
    struct edge
    {
            int to , nxt;
    } e[MAXM];
    
    template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
    template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
    template <typename T> inline void read(T &x)
    {
        T f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
    }
    inline void addedge(int u,int v)
    {
            tot++;
            e[tot] = (edge){v,head[u]};
            head[u] = tot;
    }
    
    int main()
    {
            
            read(n); read(m);
            for (int i = 1; i <= m; i++)
            {
                    int u , v;
                    read(u); read(v);
                    addedge(u,v);
            }
            for (int i = 1; i <= n; i++)
            {
                    for (int j = 1; j <= n; j++) dist[i][j] = inf;
                    queue< int > q;
                    q.push(i);        
                    dist[i][i] = 0;
                    while (!q.empty())
                    {
                            int cur = q.front();
                            q.pop();
                            for (int j = head[cur]; j; j = e[j].nxt)
                            {
                                    int v = e[j].to;
                                    if (dist[i][cur] + 1 < dist[i][v])
                                    {
                                            dist[i][v] = dist[i][cur] + 1;
                                            q.push(v);
                                    }
                            }
                    }
            }
            for (int i = 1; i <= n; i++)
            {
                    for (int j = 1; j <= n; j++)
                    {
                            if (i == j || dist[i][j] == inf) continue;
                            int pos = j;
                            for (int k = 0; k < 3; k++)
                            {
                                    if (f[i][k] == 0 || dist[i][pos] > dist[i][f[i][k]])
                                            swap(pos,f[i][k]);
                            }
                    }
                    for (int j = 1; j <= n; j++)
                    {
                            if (i == j || dist[j][i] == inf) continue;
                            int pos = j;
                            for (int k = 0; k < 3; k++)
                            {
                                    if (g[i][k] == 0 || dist[pos][i] > dist[g[i][k]][i])
                                            swap(pos,g[i][k]);
                            }
                    }
            }
            int ans = 0;
            for (int b = 1; b <= n; b++)
            {
                    for (int c = 1; c <= n; c++)
                    {
                            if (b == c || dist[b][c] == inf) continue;
                            for (int t1 = 0; t1 < 3; t1++)
                            {
                                    int a = g[b][t1];
                                    if (a == 0 || a == c) continue;
                                    for (int t2 = 0; t2 < 3; t2++)
                                    {
                                            int d = f[c][t2];
                                            if (d == 0 || d == b || d == a) continue;
                                            if (dist[a][b] + dist[b][c] + dist[c][d] > ans)
                                            {
                                                    ans = dist[a][b] + dist[b][c] + dist[c][d];
                                                    res = (info){a,b,c,d};
                                            }
                                    }    
                            }        
                    }
            }
            printf("%d %d %d %d
    ",res.a,res.b,res.c,res.d);
            
            return 0;
        
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9736720.html
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