【UVa 208】Firetruck

The Center City ?re department collaborates with the transportation department to maintain maps
of the city which re?ects the current status of the city streets. On any given day, several streets are
closed for repairs or construction. Fire?ghters need to be able to select routes from the ?restations to
?res that do not use closed streets.
Central City is divided into non-overlapping ?re districts, each containing a single ?restation. When
a ?re is reported, a central dispatcher alerts the ?restation of the district where the ?re is located and
gives a list of possible routes from the ?restation to the ?re. You must write a program that the central
dispatcher can use to generate routes from the district ?restations to the ?res.
Input
The city has a separate map for each ?re district. Streetcorners of each map are identi?ed by positive
integers less than 21, with the ?restation always on corner #1. The input ?le contains several test cases
representing di?erent ?res in di?erent districts.
• The ?rst line of a test case consists of a single integer which is the number of the streetcorner
closest to the ?re.
• The next several lines consist of pairs of positive integers separated by blanks which are the
adjacent streetcorners of open streets. (For example, if the pair 4 7 is on a line in the ?le, then
the street between streetcorners 4 and 7 is open. There are no other streetcorners between 4 and
7 on that section of the street.)
• The ?nal line of each test case consists of a pair of 0’s.
Output
For each test case, your output must identify the case by number (‘CASE 1:’, ‘CASE 2:’, etc). It must
list each route on a separate line, with the streetcorners written in the order in which they appear on
the route. And it must give the total number routes from ?restation to the ?re. Include only routes
which do not pass through any streetcorner more than once. (For obvious reasons, the ?re
department doesn’t want its trucks driving around in circles.)
Output from separate cases must appear on separate lines.
Sample Input
6
1 2
1 3
3 4
3 5
4 6
5 6
2 3
2 4
0 0
4
2 3
3 4
5 1
1 6
7 8
8 9
2 5
5 7
3 1
1 8
4 6
6 9
0 0
Sample Output
CASE 1:
1 2 3 4 6
1 2 3 5 6
1 2 4 3 5 6
1 2 4 6
1 3 2 4 6
1 3 4 6
1 3 5 6
There are 7 routes from the firestation to streetcorner 6.
CASE 2:
1 3 2 5 7 8 9 6 4
1 3 4
1 5 2 3 4
1 5 7 8 9 6 4
1 6 4
1 6 9 8 7 5 2 3 4
1 8 7 5 2 3 4
1 8 9 6 4
There are 8 routes from the firestation to streetcorner 4.

深搜即可，无需多讲。

要注意的是，要先判断起点是否和终点在一起（即在一个连通分量中）。否则会超时。这里用了传递闭包。网上说可以用并查集、TARJAN什么的。。。。

#include<cstdio>
#include<cstring>

using namespace std;

int k, path[25], len, tot;
bool v[25], g[25][25];

void solve(int x)
{
    v[x] = false;
    path[len++] = x;
    if (x == k)
    {
        tot++;
        for (int i = 0; i < len; ++i)
        {
            if (i) printf(" ");
            printf("%d", path[i]);
        }
        printf("\n");
        v[x] = true;
        len--;
        return;
    }
    for (int i = 1; i <= 21; ++i)
        if (g[x][i] && v[i])
            solve(i);
    v[x] = true;
    len--;
}

bool it_can_work()
{
    bool dis[25][25];
    memcpy(dis, g, sizeof(g));
    for (int m = 1; m <= 21; ++m)
        for (int i = 1; i <= 21; ++i)
            for (int j = 1; j <= 21; ++j)
                if (i != m && i != j && j != m)
                    dis[i][j] = dis[i][j] || (dis[i][m] && dis[m][j]);
    return dis[1][k];
}

int main()
{
    int kase = 0;
    while (scanf("%d", &k) == 1)
    {
        printf("CASE %d:\n", ++kase);
        int x, y;
        memset(v, true, sizeof(v));
        memset(g, false, sizeof(g));
        while (scanf("%d%d", &x, &y) == 2 && x && y)
            g[x][y] = g[y][x] = true;
        len = tot = 0;
        if (it_can_work())solve(1);
        printf("There are %d routes from the firestation to streetcorner %d.\n", tot, k);
    }
    return 0;
}