【prim + kruscal 】 最小產生樹模板

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來源：dlut oj1105: Zhuo’s DreamTime Limit: 1 Sec Memory Limit: 128 MB
Submit: 40 Solved: 14
[Submit][Status][Web Board]Description

Zhuo is a lovely boy and always make day dream. This afternoon he has a dream that he becomes the king of a kingdom called TwoBee.

In the dream Zhuo faced a hard situation: he should make a traffic construction plan for the kingdom. We have known that the kingdom consists of N cities and M dirt roads, it’s very uncomfortable when travelling in the road. Now he would choose some roads to rebuild to concrete road. To avoid called TwoBee by the common people, Zhuo wants to show some excellent thing to this world while it needs your help. He wants his plan achieve two goals:

1. We should choose just N-1 roads to rebuild. After rebuild we can also travel between any two cities by the concrete roads.

2. We want to minimize the longest length among the chosen roads.

So you should write a program to calculate the total length of the chosen roads.

Input

There are multiple test cases.

The first line contains two integers N and M. (1 <= N <= 100, 1 <= M <= 10000)

In the next M lines, each line contains three integers a, b, w, indicating that there is a dirt road between city a and city b with length w. (1 <= a, b <= N, 1 <= w <= 1000)

Output

For each test case, output an integer indicating the total length of the chosen roads. If there is no solution, please output -1.

Sample Input

3 3

1 2 1 1 3 1 2 3 3

Sample Output

2

HINTSource

Rainy

 

 

 

prim演算法

#include <cstdio>#include <iostream>#include <memory.h>using namespace std;const int maxn=102;const int inf=1<<30;int map[maxn][maxn];int dis[maxn];bool flag[maxn];int a,b,c,n,m,ans; void init(){    memset(map,-1,sizeof(map));    for(int i=0;i<m;i++)    {        scanf("%d%d%d",&a,&b,&c);        if(map[a][b]==-1 || c< map[a][b])        {            map[a][b]=map[b][a]=c;        }    }} void prim(){    memset(flag,false,sizeof(flag));    for(int i=1;i<=n;i++)dis[i]=inf;    dis[1]=0;    ans=0;    for(int j=1;j<=n;j++)    {        int now,value=inf;        for(int i=1;i<=n;i++)        {            if(flag[i]==false && dis[i]<value)            {                value=dis[i];                now=i;            }        }        if(value==inf)        {            cout << "-1" <<endl;            return;        }        flag[now]=true;        ans+=dis[now];        for(int i=1;i<=n;i++)        {            if(!flag[i] && map[now][i]!=-1 && dis[i]>map[now][i])            dis[i]=map[now][i];        }    }     cout << ans <<endl;} int main(){    //‘freopen("in.txt","r",stdin);    while(cin >> n >> m)    {        init();        prim();    }    return 0;}

 

 

 

 

 下面是最小產生樹的Kruskal演算法，這個演算法原理看起來很複雜，但實現起來很簡單：開始的時候是每個頂點一棵樹，並將邊按權重升序排列。然後從前到後按循序選邊，如果當前選擇的邊的兩個頂點分在兩棵不同的樹中，則將該邊加入到最小產生樹中，併合當前邊串連的兩棵樹，如果邊得兩個頂點在相同的樹中，則不做任何處理,需要注意的是這個演算法是針對無向連通圖的,如果是有限圖,則需要在演算法中做些處理,但演算法原理是一樣的。kruskal演算法： 

#include <cstdio>#include <iostream>#include <memory.h>#include <algorithm>using namespace std;const int maxn=10002;int m,n;int cnt,length;int parent[maxn]; struct node{    int from;    int to;    int val;}s[maxn]; bool cmp(const node &l,const node &r){    return l.val<r.val;} void init(){    for(int i=0;i<maxn;i++)    {        parent[i]=i;    }} int find(int x){    while(x!=parent[x])    {        x=parent[x];    }    return x;} void merge(int a,int b,int val){    int roota=find(a);    int rootb=find(b);    if(roota!=rootb)    {        cnt++;        parent[rootb]=roota;        length+=val;    }} void kruskal(){    sort(s,s+n,cmp);    for(int i=0;i<n;i++)    {        merge(s[i].from,s[i].to,s[i].val);        if(cnt==m-1)break;    }    if(cnt<m-1)    {        length=-1;    }} int main(){   // freopen("in.txt","r",stdin);    while(cin >> m >> n)    {        init();        cnt=0,length=0;        memset(s,0,sizeof(s));        for(int i=0;i<n;i++)        {              cin >> s[i].from >> s[i].to >> s[i].val;        }       kruskal();        cout << length <<endl;    }    return 0;}

 

 

【prim + kruscal 】 最小產生樹模板