【并查集最小生成树】POJ2421&&HDU1102：Constructing Roads

Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

Sample Output

题目大意：有N个村庄，编号从1到N，你应该建造一些道路，使每两个村庄可以相互连接。我们说A和B两个村庄相连，当且仅当A和B之间有一条公路，或者存在一个村庄C，这样在A和C之间有一条公路，C和B相通。
我们知道在一些村庄之间已经有一些道路，而你的工作就是修建一些道路，这样所有的村庄都连接起来，所有修建的道路的长度都是最小的。

第一行是一个整数N（3 <= N <= 100），这是村庄的数量。然后是N行，其中第i行包含N个整数，这N个整数的第j个是村i和村j之间的距离（距离应该是[1，1000]内的整数）。
然后存在整数Q（0≤Q≤N*（N + 1）/ 2）。然后进入Q线，每条线包含两个整数a和b（1 <= a <b <= N），这意味着a村和b村之间的道路已经建成。

你应该输出一行包含一个整数，这是所有要建造的道路的长度，这样所有的村庄都连接起来了，这个值是最小的。

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn = 5600;
 7 int n,cnt,q;
 8 struct node
 9 {
10     int u,v;
11     int w;
12 };
13 node edge[110];
14 int mapp[maxn][maxn],fa[maxn];
15 bool cmp(node a,node b)
16 {
17     return a.w<b.w;
18 }
19 void init()
20 {
21     for(int i=0;i<n;i++)
22     {
23         fa[i]=i;
24     }
25 }
26 int findd(int x)
27 {
28     if(fa[x]==x)
29         return x;
30     else
31         return fa[x]=findd(fa[x]);
32 }
33 void Union(int x,int y)
34 {
35     int fx=findd(x),fy=findd(y);
36     if(fx!=fy)
37     {
38         fa[fy]=fx;
39     }
40 }
41 
42 int main()
43 {
44     while(~scanf("%d",&n))
45     {
46         for(int i=1;i<=n;i++)
47         {
48             for(int j=1;j<=n;j++)
49             {
50                 scanf("%d",&mapp[i][j]);
51             }
52         }
53         int num=0;
54         for(int i=1;i<n;i++)
55         {
56             for(int j=i+1;j<=n;j++)
57             {
58                 edge[num].u=i;
59                 edge[num].v=j;
60                 edge[num].w=mapp[i][j];
61                 num++;
62             }
63         }
64         scanf("%d",&q);
65         int a,b;
66         cnt=0;
67         init();
68         for(int i=0;i<q;i++)
69         {
70             scanf("%d%d",&a,&b);
71             if(findd(a)!=findd(b))
72             {
73                 Union(a,b);
74                 cnt++;
75             }
76         }
77         sort(edge,edge+num,cmp);
78         int sum=0;
79         int u,v;
80         for(int i=0;i<num;i++)
81         {
82             u=edge[i].u;
83             v=edge[i].v;
84             if(findd(u)!=findd(v))
85             {
86                 sum+=edge[i].w;
87                 cnt++;
88                 Union(u,v);
89             }
90             if(cnt>=n-1)
91                 break;
92         }
93         cout << sum << endl;
94     }
95     return 0;
96 }

View Code

【并查集 最小生成树】POJ2421&&HDU1102：Constructing Roads

【并查集最小生成树】POJ2421&&HDU1102：Constructing Roads