POJ 2421 Constructing Roads （kruskal + 并查集）

题目链接：POJ 2421

Describe:
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:
3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output:
179

题目大意：

有n个房子，有些房子之间已经存在路，问最少修多长的路可以保证所有房子连通。

解题思路：

题目给的两个房子之间的距离就是边的权重，使边的权重最小，即最小生成树，kruskal+并查集就可以解决（感觉kruskal和并查集好像分不开）。

AC代码：

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <queue>
 5 #define rep(i,n) for(int i = 0;i < n;i++) // 从别人代码里偷学过来的
 6 using namespace std;
 7 struct edge // 定义边
 8 {
 9     int e,w,st,ed; // e表示这个边是本来就存在还是新建的，w为权重，st和ed是边的顶点
10     bool operator < (const edge &a) const // 比较函数
11     {
12         if(e != a.e) return e < a.e; // 此处注意，出现过的边优先级最高，即先比较e
13         else return w > a.w;
14     }
15 };
16 int g[100][100],d[100][100],arr[100],r[100]; // g为图，d存储距离，arr和r为并查集
17 int n,qq,x,y,ans,cnt;
18 int getfather(int x) // 以下两个函数并查集常规操作
19 {
20     return arr[x] == x?x:(arr[x] = getfather(arr[x]));
21 }
22 void mergeset(int x, int y)
23 {
24     x = getfather(x); y = getfather(y);
25     if(x != y)
26     {
27         if(r[x] <= r[y]) arr[x] = y;
28         else arr[y] = x;
29         if(r[x] == r[y] && x != y) r[y]++;
30     }
31 }
32 int main()
33 {
34     ans = 0; // 初始化答案
35     cnt = 0; // pop边时的计数器
36     memset(r,1,sizeof(r));
37     priority_queue<edge> q; // 针不戳，优先队列针不戳
38     scanf("%d",&n);
39     rep(i,n) arr[i] = i; // 并查集初始化
40     rep(i,n)
41     {
42         rep(j,n)
43         {
44             scanf("%d",&d[i][j]); // 输入距离
45         }
46     }
47     scanf("%d",&qq);
48     rep(i,qq)
49     {
50         scanf("%d%d",&x,&y); // 邻接矩阵记录边
51         x--; y--; // 注意下标
52         g[x][y] = g[y][x] = 1;
53     }
54     rep(i,n)
55     {
56         for(int j = i+1; j < n; j++) // 画个图就可以看出来
57         {
58             edge tmp; // 插入边
59             tmp.w = d[i][j];
60             if(g[i][j] == 1) tmp.e = 1; // 判断是否为已经存在的边
61             else tmp.e = 0;
62             tmp.st = i;
63             tmp.ed = j;
64             q.push(tmp); // 入对
65         }
66     }
67     while(cnt != (n-1)) // 生成树有n-1条边
68     {
69         edge tmp = q.top();
70         q.pop();
71         int t1,t2;
72         t1 = getfather(tmp.st);
73         t2 = getfather(tmp.ed);
74         if(t1 == t2) continue; // 判断是否已经成一棵树
75         mergeset(tmp.st,tmp.ed); // 不是一棵树则合并
76         if(g[tmp.st][tmp.ed] == 0) ans += tmp.w; // 注意要加上新建的边
77         cnt++;
78     }
79     printf("%d",ans);
80     return 0;
81 }

 

posted @ 2020-08-18 17:22  不敢说的梦  阅读(126)  评论(0)    收藏  举报