HDOJ---3371 Connect the Cities[kruskal算法]-优快云博客

本文探讨了在海平面上升导致城市被淹没的情况下，如何通过算法建立城市之间的连接，最小化建设成本。详细介绍了输入输出格式、案例分析及算法实现过程。

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4439 Accepted Submission(s): 1298

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

Author

dandelion

Source

HDOJ Monthly Contest – 2010.04.04

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lcy

并查集的记忆化搜索

code:

 1 #include <iostream>   
 2 #include <iomanip>   
 3 #include <fstream>   
 4 #include <sstream>   
 5 #include <algorithm>   
 6 #include <string>   
 7 #include <set>   
 8 #include <utility>   
 9 #include <queue>   
10 #include <stack>   
11 #include <list>   
12 #include <vector>   
13 #include <cstdio>   
14 #include <cstdlib>   
15 #include <cstring>   
16 #include <cmath>   
17 #include <ctime>   
18 #include <ctype.h> 
19 using namespace std;
20 const int N=1000000;
21 int Tc,n,m,k,done,ans;
22 int father[N];
23 int num[N];
24 
25 int findset(int p){
26     if (father[p]==p)
27         return p;
28     father[p]=findset(father[p]);
29     return father[p];
30 }
31 
32 typedef struct edge
33 {
34     int x,y;
35     int len;
36 }Edge;
37 Edge edge[N];
38 
39 int cmp(const Edge &a,const Edge &b){
40     return a.len<b.len;
41 }
42 
43 void kruskal(){
44     if (done==n-1) return;
45     sort(edge+1,edge+m+1,cmp);
46     for (int i=1;i<=m;i++){
47         int x=edge[i].x,y=edge[i].y;
48         if (findset(x)!=findset(y)){
49             if (num[father[x]]<num[father[y]]){
50                 num[father[y]]+=num[father[x]];
51                 father[father[x]]=father[y];
52             }
53             else{
54                 num[father[x]]+=num[father[y]];
55                 father[father[y]]=father[x];
56             }
57             done++;
58             ans+=edge[i].len;
59             if (done==n-1) return;
60         }
61     }
62 }
63 
64 int main(){
65     scanf("%d",&Tc);
66     for (int q=1;q<=Tc;q++)
67     {
68         done=0; ans=0;
69         int i,j,t,one,x;
70         scanf("%d%d%d",&n,&m,&k);
71         for (i=1;i<=n;i++){
72             father[i]=i;
73             num[i]=1;
74         }
75         for (i=1;i<=m;i++)
76             scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].len);
77         for (i=1;i<=k;i++){
78             scanf("%d%d",&t,&one);
79             for (j=1;j<t;j++){
80                 scanf("%d",&x);
81                 if (findset(x)!=findset(one)){
82                     if (num[father[x]]<num[father[one]]){
83                         num[father[one]]+=num[father[x]];
84                         father[father[x]]=father[one];
85                     }
86                     else{
87                         num[father[x]]+=num[father[one]];
88                         father[father[one]]=father[x];
89                     }
90                     done++;
91                 }
92             }
93         }
94         kruskal();
95         if (done==n-1) printf("%d\n",ans);
96         else printf("-1\n");
97     }
98 }

还可以水过去

code:

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 int f[505],L[505];
 5 struct node
 6 {
 7     int x,y,s;
 8 }dd[25005];
 9 bool cmp(node a,node b)
10 {
11     return a.s<b.s;
12 }
13 int find(int a)
14 {
15     if(a!=f[a])
16         a=find(f[f[a]]);
17     return a;
18 }
19 void Union(int x,int y)
20 {
21     if(x>y)
22         f[x]=y;
23     else
24         f[y]=x;
25 }
26 int main()
27 {
28     int n,m,k,t;
29     int i,j;
30     int T;
31     scanf("%d",&T);
32     while(T--)
33     {
34         scanf("%d%d%d",&n,&m,&k);
35         for(i=0;i<m;i++)
36             scanf("%d %d %d",&dd[i].x,&dd[i].y,&dd[i].s);
37         sort(dd,dd+m,cmp);
38         for(i=1;i<=n;i++)
39             f[i]=i;
40         int One,Two;
41         for(i=0;i<k;i++)
42         {
43             scanf("%d%d",&t,&One);
44             for(j=1;j<t;j++)
45             {
46                 scanf("%d",&Two);
47                 int xx=find(One);
48                 int yy=find(Two);
49                 if(xx!=yy)
50                     Union(xx,yy);
51             }
52         }
53         int sum=0;
54         for(i=0;i<m;i++)
55         {
56             int xx=find(dd[i].x);
57             int yy=find(dd[i].y);
58             if(xx!=yy)
59             {
60                 Union(xx,yy);
61                 sum+=dd[i].s;
62             }
63         }
64         int cnt=0;
65         for(i=1;i<=n;i++)
66             if(i==f[i])
67                 cnt++;
68         if(cnt>1)
69             printf("-1\n");
70         else
71             printf("%d\n",sum);
72     }
73     return 0;
74 }

以树的数量进行记录

code:

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<cstdlib>
  4 struct node
  5 {
  6     int x,y,z;
  7 }arr[25100];
  8 int n,m,k,t,a[510],pre[510],cnt;
  9 bool v[510];
 10 int cmp(const void *p,const void *q)
 11 {
 12     return ((node *)p)->z-((node *)q)->z;
 13 }
 14 int find(int x)
 15 {
 16     while(x!=pre[x])
 17         x=pre[x];
 18     return x;
 19 }
 20 void bcj()
 21 {
 22     int ans=0,flag=0,cat=0;
 23     bool vit[510],v[510];
 24     for(int i=1;i<=n;i++)
 25         vit[i]=false;
 26     for(int i=0;i<m;i++)
 27     {
 28         int fx=find(arr[i].x);
 29         int fy=find(arr[i].y);
 30         if(fx!=fy)
 31         {
 32             pre[fx]=fy;
 33             cat++;
 34             ans+=arr[i].z;
 35         }
 36         if(cat==cnt-1)
 37             break;
 38     }
 39     for(int j=1;j<=n;j++)
 40         vit[find(j)]=true;
 41     for(int j=1;j<=n;j++)
 42     {
 43         if(vit[j])
 44         {
 45             flag++;
 46             vit[j]=false;
 47         }
 48     }
 49     if(flag==1)
 50     {
 51         printf("%d\n",ans);
 52         return;
 53     }
 54     else
 55     {
 56         printf("-1\n");
 57         return;
 58     }
 59 }
 60 int main()
 61 {
 62     int p;
 63     while(~scanf("%d",&p))
 64     {
 65         while(p--)
 66         {
 67             cnt=0;
 68             scanf("%d%d%d",&n,&m,&k);
 69             for(int i=1;i<=n;i++)
 70             {
 71                 pre[i]=i;v[i]=false;
 72             }
 73             for(int i=0;i<m;i++)
 74                 scanf("%d%d%d",&arr[i].x,&arr[i].y,&arr[i].z);
 75             qsort(arr,m,sizeof(arr[0]),cmp);
 76             for(int i=0;i<k;i++)
 77             {
 78                 scanf("%d",&t);
 79                 for(int j=1;j<=t;j++)
 80                     scanf("%d",&a[j]);
 81                 for(int j=1;j<t;j++)
 82                 {
 83                     int fx=find(a[j]);
 84                     int fy=find(a[j+1]);
 85                     if(fx!=fy)
 86                         pre[fx]=fy;
 87                 }
 88             }
 89             for(int i=1;i<=n;i++)
 90             {
 91                 int fx=find(i);
 92                 if(!v[fx])
 93                 {
 94                     v[fx]=true;
 95                     cnt++;
 96                 }
 97             }
 98             //printf("cnt=%d\n",cnt);
 99             bcj();
100         }
101     }
102     return 0;
103 }

注意：

刚开始的时候，findset()这个函数一直是不对，以后要改过来！

正确写法：

1 int findset(int p){
2     if (father[p]==p)
3         return p;
4     father[p]=findset(father[p]);
5     return father[p];
6 }