UVA208-Firetruck(并查集+dfs)

Posted 2021-01-03 npugen

Problem UVA208-Firetruck

Accept：1733  Submit：14538

Time Limit: 3000 mSec

 Problem Description

 

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 Ouput

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.

 

题解：水题，寻找路径之前先判断是否连通，并查集可以搞定，搜索时因为要保证字典序，因此先从标号小的开始搜。

 

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int maxn = 25;
 6 
 7 vector< vector<int> > G(maxn);
 8 int tar,ans;
 9 int pre[maxn];
10 bool vis[maxn];
11 int sta[maxn];
12 
13 int findn(int x){
14     return x == pre[x] ? x : pre[x] = findn(pre[x]);
15 }
16 
17 void merge_node(int x,int y){
18     int fx = findn(x);
19     int fy = findn(y);
20     if(fx != fy){
21         pre[fx] = fy;
22     }
23 }
24 
25 void dfs(int u,int pos){
26     sta[pos] = u;
27     if(u == tar){
28         ans++;
29         printf("%d",sta[1]);
30         for(int i = 2;i <= pos;i++){
31             printf(" %d",sta[i]);
32         }
33         printf("
");
34         return;
35     }
36     for(int i = 0;i < G[u].size();i++){
37         int v = G[u][i];
38         if(!vis[v]){
39             vis[v] = true;
40             dfs(v,pos+1);
41             vis[v] = false;
42         }
43     }
44 }
45 
46 int iCase = 1;
47 
48 int main()
49 {
50 #ifdef GEH
51     freopen("helloworld.01.inp","r",stdin);
52 #endif
53     while(~scanf("%d",&tar)){
54         int u,v;
55         for(int i = 0;i < maxn;i++){
56             G[i].clear();
57             pre[i] = i;
58         }
59         while(~scanf("%d%d",&u,&v) && (u||v)){
60             G[u].push_back(v);
61             G[v].push_back(u);
62             merge_node(u,v);
63         }
64         ans = 0;
65         printf("CASE %d:
",iCase++);
66         if(findn(1) != findn(tar)){
67             printf("There are %d routes from the firestation to streetcorner %d.
",0,tar);
68         }
69         else{
70             for(int i = 0;i < maxn;i++){
71                 if(G[i].size()){
72                     sort(G[i].begin(),G[i].end());
73                 }
74             }
75             memset(vis,false,sizeof(vis));
76             vis[1] = true;
77             dfs(1,1);
78             printf("There are %d routes from the firestation to streetcorner %d.
",ans,tar);
79         }
80     }
81     return 0;
82 }