HDOJ 1224 Free DIY Tour （Floyd + 打印路径）

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本题关键是打印路径，加了一个path数组。path[i][j]存放的是从i到j的路径中，i的下一个点。在初始化path的时候，所有已知的通路都是从点x到点y，那么path[x][y]=x。注意看一下路径保存和打印的程序段。

#include<cstdio>#include<cstring>const int INF = 0xfffffff;int main(){    int t, cnt = 0, n, m, u, v, x, y;    int fun[105];    int a[105][105], path[105][105];    scanf("%d", &t);    while (t--)    {        cnt++;        if (cnt != 1) printf("\n");        scanf("%d", &n);        for (int i = 1; i <= n; i++) scanf("%d", &fun[i]);        fun[n+1] = 0;        scanf("%d", &m);        for (int i = 1; i <= n+1; i++)            for (int j = 1; j <= n+1; j++)            a[i][j] = -INF;        memset(path, -1, sizeof(path));        for (int i = 0; i < m; i++)        {            scanf("%d%d", &x, &y);            a[x][y] = fun[y]; // 从x到y会在原有的兴趣值上加上y点的兴趣值            path[x][y] = y;        }        for (int k = 1; k <= n+1; k++)            for (int i = 1; i <= n+1; i++)                for (int j = 1; j <= n+1; j++)                if (a[i][k] != -INF && a[k][j] != -INF && a[i][k]+a[k][j] > a[i][j])                {                    a[i][j] = a[i][k]+a[k][j];                    path[i][j] = path[i][k];                }        u = 1;        v = n+1;        printf("CASE %d#\n", cnt);        printf("points : %d\n", a[u][v]);        printf("circuit : ");        while (u != v)        {            printf("%d->", u);            u = path[u][v];        }        printf("%d\n", 1);    }}

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