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hdu 3371 Connect the Cities

2024-09-07 09:28:12   216人阅读

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3371

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18733    Accepted Submission(s): 4563


Problem Description
In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.  
 

 

Input
The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
 

 

Output
For each case, output the least money you need to take, if it’s impossible, just output -1.
 

 

Sample Input
1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6
 

 

Sample Output
1
 
题目大意：有n个城市，因为海平面上升，毁坏了一些城市之间的道路，需要重新修建一些道路保持城市之间的畅通。
　　　　　给出一些可以选择的道路以及花费，然后再给出一些依旧完好无损的道路。问是否能够保持所有城市之间的
　　　　　畅通，以及最少花费是多少。
 
解题思路：用kruskal算法解决，建图的时候注意完好无损的路就好。
 
AC代码：






20022026
2017-03-06 19:52:45
Accepted
3371
920MS
1696K
1381 B
G++







 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 struct point 
 8 {
 9     int x,y,l;
10 }p[25010];
11 int parent[510],n,m,k;
12 bool cmp(point a, point b)
13 {
14     return a.l < b.l;
15 }
16 int find(int x)
17 {
18     int s,tmp;
19     for (s = x; parent[s] >= 0; s = parent[s]);
20     while (s != x)
21     {
22         tmp = parent[x];
23         parent[x] = s;
24         x = tmp;
25     }
26     return s;
27 }
28 void Union(int A, int B)
29 {
30     int a = find(A), b = find(B);
31     int tmp = parent[a] + parent[b];
32     if (parent[a] < parent[b])
33     {
34         parent[b] = a;
35         parent[a] = tmp;
36     }
37     else
38     {
39         parent[a] = b;
40         parent[b] = tmp;
41     }
42 }
43 void kruskal()
44 {
45     int i,u,v,sum = 0,j = 0;
46     for (i = 1; i <= n; i ++)
47         parent[i] = -1;
48     for (i = 0; i < m; i ++)
49     {
50         u = p[i].x; v = p[i].y;
51         if (find(u) != find(v))
52         {
53             sum += p[i].l;
54             Union(u,v);
55             j ++;
56         }
57         if (j == n-1) break;
58     }
59     if (j == n-1)
60     printf("%d\n",sum);
61     else
62     printf("-1\n");
63 }
64 int main ()
65 {
66     int i,j,t,d;
67     int f[510];
68     scanf("%d",&t);
69     while (t --)
70     {
71         scanf("%d%d%d",&n,&m,&k);
72         for (i = 0; i < m; i ++)
73             scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].l);
74         while (k --)
75         {
76             scanf("%d",&d);
77             for (i = 0; i < d; i ++)
78             scanf("%d",&f[i]);
79             for (i = 0; i < d-1; i ++)
80             {
81                 p[m].x = f[i];
82                 p[m].y = f[i+1];
83                 p[m].l = 0;
84                 m ++;
85             }
86         }
87         sort(p,p+m,cmp);        
88         kruskal();
89     }
90     return 0;
91 }




 





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