《算法竞赛进阶指南》---- 观光之旅

 

 floyd算法

#include <bits/stdc++.h>
using namespace std;

#define io std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define x first
#define y second
#define fr front
#define db double
//int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

//int dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
//int dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};

//int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
//int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};

typedef pair<int, int> PII;
typedef long long LL ;

const int N = 110, INF = 0x3f3f3f3f;
                                                    // 
int g[N][N], d[N][N]; // g记录原始的 2 点之间的距离( A -> K , K -> B ), d 记录  更新之后 的 A -> B
int p[N][N];
int path[N], cnt;
int n, m;

void get_path(int x, int y)
{
    if (p[x][y] == 0) return ;
    
    get_path( x, p[x][y]);
    path[cnt ++ ] = p[x][y];
    get_path(p[x][y], y);
}

int main()
{
	io; 
    cin >> n >> m;
    memset(g, 0x3f, sizeof g);
    for (int i = 1;i <= n;i ++ ) g[i][i] = 0;
    
    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = min(g[a][b], c);
    }
    
    int ans = INF;
    
    memcpy(d, g, sizeof g); //  g 留做备份
    for (int k = 1;k <= n;k ++ ) // 设 k 为 路线中节点的最大编号
    {
        for (int i = 1;i < k;i ++ )
       {
        //   cout << ans << endl;
           for (int j = i + 1;j < k;j ++ )
           {
               //cout << ans << endl;
             //d[j][i]则表示j到i且经过的节点编号小于k,因为在环中k就是最大的，只能经过小于k的节点了;
            if ((LL)g[i][k] + g[k][j] + d[i][j] < ans) // g[i][k] 与 g[k][j] 是两点之间的距离 题目要求环 点数 >= 3
            {
                // i -> k -> j - > i 
                //cout << ans << endl;
                ans = g[i][k] + g[k][j] + d[i][j];
        
                cnt = 0;
                path[cnt ++ ] = k;
                path[cnt ++ ] = i;

                get_path(i, j); 
                
                path[cnt++] = j;
            }
           }
       } 
        for (int i = 1;i <= n;i ++ )
        for (int j = 1;j <= n;j ++ )
            if (d[i][j] > d[i][k] + d[k][j])
            {
                d[i][j] = d[i][k] + d[k][j];
                p[i][j] = k; // 由 k 点转移
            }
    }
    
    if (ans == INF) cout << "No solution.";
    else for (int i = 0;i < cnt;i ++ ) cout << path[i] << ' ';
    
	return 0;
}