POJ 1734 Sightseeing trip (Floyd 最小环＋记录路径)

#include <cstdio>
#include <cstring>
int const MAX = 105;
int const INF = 0xfffffff;
int d[MAX][MAX], map[MAX][MAX], path[MAX][MAX];
int n, m, ans[MAX], mi, cnt;

void init()
{
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            d[i][j] = INF;
            map[i][j] = INF;
            path[i][j] = i;
        }
    }
}

void Floyd()
{
    mi = INF;
    for(int k = 1; k <= n; k++)
    {
        //求环
        for(int i = 1; i < k; i++)
        {
            for(int j = 1; j < i; j++)
            {
                if(d[i][j] + map[i][k] + map[k][j] < mi)
                {
                    //此时的k值不在d[i][j]的路径中，所以找到一个环
                    mi = d[i][j] + map[i][k] + map[k][j];
                    //记录路径把j当起点顺序是: j -> j到i最短路的路径 -> i -> k
                    int tmp = j;
                    cnt = 0;
                    while(tmp != i)
                    {
                        ans[cnt++] = tmp;
                        tmp = path[i][tmp];
                    }
                    ans[cnt++] = i;
                    ans[cnt++] = k;
                }
            }
        }
        //求最短路
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                if(d[i][k] + d[k][j] < d[i][j])
                {
                    d[i][j] = d[i][k] + d[k][j];
                    //path[i][j]表示i到j最短路径上j前面的一个点 
                    //所以此时i到j的最短路径上j前一个点为k到j最短路径上j的前一个点
                    path[i][j] = path[k][j]; 
                }
            }
        }
    }
}

int main()
{
    int u, v, w;
    scanf("%d %d", &n, &m);
    init();
    for(int i = 0; i < m; i++)
    {
        scanf("%d %d %d", &u, &v, &w);
        if(d[u][v] > w)  //找最小环遇到平行边时取其最小边权
        {
            d[u][v] = d[v][u] = w;
            map[u][v] = map[v][u] = w;
        }
    }
    Floyd();
    if(mi == INF)
        printf("No solution.\n");
    else
    {
        printf("%d", ans[0]);
        for(int i = 1; i < cnt; i++)
            printf(" %d", ans[i]);
        printf("\n");
    }
}