poj 1258 Agri-Net 最小生成树

最新推荐文章于 2019-08-24 23:34:37 发布

原创最新推荐文章于 2019-08-24 23:34:37 发布 · 382 阅读

0 ·

CC 4.0 BY-SA版权

算法竞赛同时被 3 个专栏收录

180 篇文章

订阅专栏

算法竞赛题解

115 篇文章

订阅专栏

图论

32 篇文章

订阅专栏

本文介绍了两种求解最小生成树的经典算法：Prim算法和Kruskal算法，并提供了完整的C++实现代码。Prim算法适用于密集图，通过不断选择与已构建部分连接且权重最小的边来逐步构建最小生成树；Kruskal算法则按边的权重从小到大排序，依次加入不会形成环的边。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem:
给了一个矩阵，求这个矩阵的最小生成树。
Solution:
密集图利用prim算法。

#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<vector>
#include<fstream>
#include<list>
using namespace std;

#define ms(s) memset(s,0,sizeof(s))
typedef unsigned long long ULL;
typedef long long LL;

const double PI = 3.141592653589;
const int INF = 0x3fffffff;

const int maxn = 105;
int G[maxn][maxn];
bool vis[maxn];

int prim(int n){//index from 1
    int mins, vtx, weight = 0;
    memset(vis, 0, sizeof(vis));
    vis[1] = true;

    for(int i = 2; i <= n; ++i){
        mins = INF;
        for(int j = 1; j <= n; ++j){
            if(!vis[j] && G[1][j]<mins){
                mins = G[1][j];
                vtx = j;
            }
        }
        weight += mins;
        vis[vtx] = true;
        for(int j = 1; j <= n; ++j){
            if(!vis[j] && G[vtx][j]<G[1][j]){
                G[1][j] = G[vtx][j];
            }
        }
    }
    return weight;
}

int main(){
//    freopen("E:\\input.txt","r",stdin);
//    freopen("/home/really/Document/output","w",stdout);
//    ios::sync_with_stdio(false);

    int n;
    int ans;
    while(~scanf("%d",&n)){
        ans = 0;
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= n; ++j){
                scanf("%d",&G[i][j]);
            }
        }
        ans = prim(n);
        printf("%d\n",ans);
    }

    return 0;
}

//Kruskal
#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<vector>
#include<fstream>
#include<list>
using namespace std;

#define ms(s) memset(s,0,sizeof(s))
typedef unsigned long long ULL;
typedef long long LL;

const double PI = 3.141592653589;
const int INF = 0x3fffffff;

const int maxn = 110;

struct Edge{
    int s, e, w;
    Edge(int ns, int ne, int nw) : s(ns), e(ne), w(nw){}
    Edge(){}
    bool operator < (const Edge& rhs) const{
        return w < rhs.w;
    }
};

int par[maxn];
void init_par(int n){
    for(int i = 0; i <= n; ++i)
        par[i] = i;
}
int getPar(int v){
    if(par[v] != v)
        par[v] = getPar(par[v]);
    return par[v];
}
void merges(int a, int b){
    par[getPar(a)] = getPar(b);
}

vector<Edge> edges;
int kruskal(int n){
    int num = 1, ans = 0;
    init_par(n);
    sort(edges.begin(), edges.end());
    for(int i = 0; i < edges.size(); ++i){
        if(getPar(edges[i].s) != getPar(edges[i].e)){
            merges(edges[i].s, edges[i].e);
            ++num;  ans += edges[i].w;
        }
        if(num == n)  break;
    }
    return ans;
}

int main(){
//    freopen("E:\\input.txt","r",stdin);
//    freopen("/home/really/Document/output","w",stdout);
//    ios::sync_with_stdio(false);

    int n, ans, w;
    while(~scanf("%d",&n)){
        ans = 0;  edges.clear();

        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= n; ++j){
                scanf("%d", &w);
                edges.push_back(Edge(i, j, w));
            }
        }

        ans = kruskal(n);
        printf("%d\n",ans);
    }

    return 0;
}