最近有几位同学催我更新，于是来摸摸鱼，来讲一下最小生成树问题。

　　所谓最小生成树（MST），就是在一张无向带权图中的一棵经过所有节点，边权和最小的一棵树。在实际生活中，可以运用于城镇之间的修路上。

　　对于MST问题，通常有两种算法，prim算法以及kruskal算法，其中最常用的算法为kruskal算法，其与并查集结合之后可以以O（ElogE）的时间复杂度解决问题，对于稀疏图而言，是一种非常优秀的算法，但是，对于稠密图而言，更适合使用prim算法。这次就只介绍一下kruskal算法。

　　kruskal算法本质是一种贪心算法，我们知道，树一定无环，我们先对所有边以边权从小到大进行排序，然后从小到大逐个选边，将端点存入并查集中，若边的两个端点发现在同一个并查集中，则会成环，就跳过这个边，否则就把这条边存入MST中。这样我们就得到了一棵最小生成树。易证这种算法的正确性。至于并查集的介绍可以参见我之前的博客。

　　例题1 hdu1233

还是畅通工程

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 83393    Accepted Submission(s): 37809

Problem Description
某省调查乡村交通状况，得到的统计表中列出了任意两村庄间的距离。省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可），并要求铺设的公路总长度为最小。请计算最小的公路总长度。
Input
测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( < 100 )；随后的N(N-1)/2行对应村庄间的距离，每行给出一对正整数，分别是两个村庄的编号，以及此两村庄间的距离。为简单起见，村庄从1到N编号。
当N为0时，输入结束，该用例不被处理。
Output
对每个测试用例，在1行里输出最小的公路总长度。
Sample Input
3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0
Sample Output
3
5
　　最小生成树模板题，代码如下
#include<map>

#include<cmath>

#include<queue>

#include<cstdio>

#include<vector>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

typedef long long ll;

typedef pair<int,int>pii;

const int maxn=(1e5)+5;

const int maxm=(1e5)+5;

vector<int>g[maxn];

int par[maxn];

int depth[maxn];

int n,m;

//并查集

void init(int n){

    for(int i=0;i<=n;i++){

        par[i]=i;

        depth[i]=0;

    }

}

int find(int x){

    if(par[x]==x)return x;

    else return par[x]=find(par[x]);

}

void unit(int x,int y){

    x=find(x);

    y=find(y);

    if(x==y)return;

    if(depth[x]<depth[y])par[x]=y;

    else{

        par[y]=x;

        if(depth[x]==depth[y])depth[x]++;

    }

}

bool same(int x,int y){

    return find(x)==find(y);

}

//kruskal算法

struct Edge{

    int u,v,w;

}edge[maxm];

bool cmp(Edge x,Edge y){

    return x.w<y.w;

}

int kruskal(){

    int ans=0;

    init(n);

    sort(edge+1,edge+m+1,cmp);

    for(int i=1;i<=m;i++){

        int a=find(edge[i].u);

        int b=find(edge[i].v);

        if(a==b)continue;//成环

        unit(a,b);

        ans+=edge[i].w;

    }

    return ans;

}

int main(){

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cin>>n;

    while(n){

        m=n*(n-1)/2;

        for(int i=1;i<=m;i++){

            cin>>edge[i].u>>edge[i].v>>edge[i].w;

        }

        cout<<kruskal()<<endl;

        cin>>n;

    }

    return 0;

}

　　例题2 poj2421

Constructing Roads
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 35679   Accepted: 16044

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

3

0 990 692

990 0 179

692 179 0

1

1 2

Sample Output

　179

题目大意：已知有某些边已经建立了，问使这张图联通的代价是多少。

解析：依然是最小生成树问题，只是先预存边到并查集中，再跑kruskal即可。

代码如下

#include<map>

#include<cmath>

#include<queue>

#include<cstdio>

#include<vector>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

typedef long long ll;

typedef pair<int,int>pii;

const int maxn=105;

const int maxm=10005;

int par[maxn];

int depth[maxn];

int n,m;

void init(int n){

    for(int i=0;i<=n;i++){

        par[i]=i;

        depth[i]=0;

    }

}

int find(int x){

    if(par[x]==x)return x;

    else return par[x]=find(par[x]);

}

void unit(int x,int y){

    x=find(x);

    y=find(y);

    if(x==y)return;

    if(depth[x]<depth[y])par[x]=y;

    else{

        par[y]=x;

        if(depth[x]==depth[y])depth[x]++;

    }

}

bool same(int x,int y){

    return find(x)==find(y);

}

struct Edge{

    int u,v,w;

}edge[maxm];

bool cmp(Edge x,Edge y){

    return x.w<y.w;

}

int main(){

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cin>>n;

    int cnt=0;

    for(int i=1;i<=n;i++){

        for(int j=1;j<=n;j++){

            int x;

            cin>>x;

            edge[++cnt].u=i;

            edge[cnt].v=j;

            edge[cnt].w=x;

        }

    }

    m=cnt;

    init(n);

    int q;

    cin>>q;

    while(q--){

        int x,y;

        cin>>x>>y;

        if(!same(x,y))unit(x,y);

    }

    sort(edge+1,edge+m+1,cmp);

    int ans=0;

    for(int i=1;i<=m;i++){

        int a=find(edge[i].u);

        int b=find(edge[i].v);

        if(a==b)continue;

        unit(a,b);

        ans+=edge[i].w;

    }

    cout<<ans<<endl;

    return 0;

}

例题3：poj2377

Bad Cowtractors
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 21954   Accepted: 8892

Description

Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn't even want to pay Bessie.

Realizing Farmer John will not pay her, Bessie decides to do the worst job possible. She must decide on a set of connections to install so that (i) the total cost of these connections is as large as possible, (ii) all the barns are connected together (so that it is possible to reach any barn from any other barn via a path of installed connections), and (iii) so that there are no cycles among the connections (which Farmer John would easily be able to detect). Conditions (ii) and (iii) ensure that the final set of connections will look like a "tree".

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..M+1: Each line contains three space-separated integers A, B, and C that describe a connection route between barns A and B of cost C.

Output

* Line 1: A single integer, containing the price of the most expensive tree connecting all the barns. If it is not possible to connect all the barns, output -1.

Sample Input

5 8

1 2 3

1 3 7

2 3 10

2 4 4

2 5 8

3 4 6

3 5 2

4 5 17

Sample Output

42

Hint

OUTPUT DETAILS:

The most expensive tree has cost 17 + 8 + 10 + 7 = 42. It uses the following connections: 4 to 5, 2 to 5, 2 to 3, and 1 to 3.
　　这题是最大生成树，只要改为从大到小排序即可，最后再并查集检查一下连通性即可。
　　代码如下
#include<map>

#include<cmath>

#include<queue>

#include<cstdio>

#include<vector>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

typedef long long ll;

typedef pair<int,int>pii;

const int maxn=(1e3)+5;

const int maxm=(2e4)+5;

vector<int>g[maxn];

int n,m;

int par[maxn];

int depth[maxn];

void init(int n){

    for(int i=0;i<=n;i++){

        par[i]=i;

        depth[i]=0;

    }

}

int find(int x){

    if(par[x]==x)return x;

    else return par[x]=find(par[x]);

}

void unit(int x,int y){

    x=find(x);

    y=find(y);

    if(x==y)return;

    if(depth[x]<depth[y])par[x]=y;

    else{

        par[y]=x;

        if(depth[x]==depth[y])depth[x]++;

    }

}

bool same(int x,int y){

    return find(x)==find(y);

}

struct Edge{

    int u,v,w;

}edge[maxm];

bool cmp(Edge x,Edge y){

    return x.w>y.w;

}

int main(){

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cin>>n>>m;

    for(int i=1;i<=m;i++){

        cin>>edge[i].u>>edge[i].v>>edge[i].w;

    }

    sort(edge+1,edge+m+1,cmp);

    init(n);

    ll ans=0;

    for(int i=1;i<=m;i++){

        int a=find(edge[i].u);

        int b=find(edge[i].v);

        if(a==b)continue;

        unit(a,b);

        ans+=edge[i].w;

    }

    int x=find(1);

    for(int i=1;i<=n;i++){

        if(find(i)!=x){

            ans=-1;

            break;

        }

    }

    cout<<ans<<endl;

    return 0;

}

总结：kruskal代码看似较长，实际上思路就是简单的贪心以及与并查集的结合，多数可以用模板直接套，修改以下细节即可。

摸结束，谢谢阅读。

最新文章

  1. android.util.TypedValue.applyDimension
  2. Android应用内语言切换实现(转)
  3. angular.js初探
  4. shell脚本ssh自动登陆服务器
  5. oracle建立表空间
  6. iOS开发 - 不进入待机(屏幕保持唤醒)---UIApplication学习
  7. 【从零开始】用node搭建一个jsonp&amp;json服务
  8. C++多态？
  9. Web缓存（一） - HTTP协议缓存
  10. 最近公共祖先（LCT）
  11. STP(Spanning Tree Protocol)
  12. Nginx http反向代理流程Proxy_pass模块
  13. top 分析
  14. Springboot2.x 集成jsp
  15. jxl和POI的区别
  16. css3实现文本渐变
  17. 入门智能家居，从 IFTTT 到 HomeKit 自动化（二）
  18. ci 3.0 默认路由放在子文件夹 无法访问的解决办法
  19. 盒子模型之margin相关技巧！
  20. mybatis模糊查询防止SQL注入

热门文章

  1. MQTT X 1.9.1 发布：资源消耗降低 80%，稳定性大幅提升
  2. a[i]之和小于N的含义
  3. Seata安装与使用
  4. wpf DataGrid相关总结
  5. R7-1 字符排队
  6. 10 个常用的 JS 工具库，80% 的项目都在用！
  7. redis的持久化方案RDB和AOF
  8. 运用python中装饰器方法来解决工作中为原有代码添加功能问题
  9. C++杂乱
  10. pat题目整理