【UVa】208 - Firetruck

Problem here

Problem

The Center City fire department collaborates with the transportation department to maintain maps
of the city which reflects the current status of the city streets. On any given day, several streets are
closed for repairs or construction. Firefighters need to be able to select routes from the firestations to
fires that do not use closed streets.
Central City is divided into non-overlapping fire districts, each containing a single firestation. When
a fire is reported, a central dispatcher alerts the firestation of the district where the fire is located and
gives a list of possible routes from the firestation to the fire. You must write a program that the central
dispatcher can use to generate routes from the district firestations to the fires.

INPUT

The city has a separate map for each fire district. Streetcorners of each map are identified by positive
integers less than 21, with the firestation always on corner #1. The input file contains several test cases
representing different fires in different districts.
• The first line of a test case consists of a single integer which is the number of the streetcorner
closest to the fire.
• 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 file, 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 final 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 firestation to the fire. Include only routes
which do not pass through any streetcorner more than once. (For obvious reasons, the fire
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

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.

Solution

Floyd 判斷b能否到達終點
再判斷a 能否到達b

#include <iostream>
#include <memory.h>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;

int a[30][30], b[30][30];
int visit[30], ans[30];
int m, cas, cnt, n;

void dfs(int c, int p){
    if(p == n){
        cnt++;
        for(int i = 0; i < c; i++){
            if(i > 0)   
                cout << " ";
            cout << ans[i];
        }
        cout << endl;
        return ;
    }

    for(int i = 1; i <= m; i++){
        if(visit[i])
            continue;
        if(!a[p][i])
            continue;
        if(!b[i][n])
            continue;
        visit[i] = 1;
        ans[c] = i;
        dfs(c+1,i);
        visit[i] = 0;
    }
}

int main(){
    cas = 0;
    while(cin >> n){
        int x, y;
        cas++;
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        memset(visit, 0, sizeof(visit));
        memset(ans, 0, sizeof(ans));
        while(cin >> x >> y){
            if(x == 0 && y == 0)
                break;

            a[x][y] = a[y][x] = 1;
            b[x][y] = b[y][x] = 1;
            m = max(m, max(x, y));
        }

        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= m; j++)
                for(int k = 1; k <= m; k++)
                    if(b[j][i] && b[i][k])
                        b[j][k] = 1;

        cnt = 0;
        visit[1] = 1;
        ans[0] = 1;
        cout << "CASE " << cas << ":" << endl;
        dfs(1, 1);
        cout << "There are " << cnt << " routes from the firestation to streetcorner " << n << "." << endl;

    }

    return 0;
}