51Nod 1212无向图最小生成树

 

prim

#include<stdio.h>
#include<string.h>
#define inf 0x3f3f3f3f
int G[1001][1001];
int vis[1001],lowc[1001];
int prim(int G[][1001],int n){
    int i,j,p,minc,res=0;
    memset(vis,0,sizeof(vis));//全部初值为0表示没有访问过；
    vis[1]=1;
    for(i=2;i<=n;i++)
        lowc[i]=G[1][i];
    for(i=2;i<=n;i++){
        minc=inf;
        p=-1;
        for(j=1;j<=n;j++){
            if(vis[j]==0&&lowc[j]<minc)
                {minc=lowc[j];p=j;}
        }
        if(inf==minc) return -1;//原图不连通
        res+=minc;
        vis[p]=1;
        for(j=1;j<=n;j++){//更新lowc[]
            if(vis[j]==0&&lowc[j]>G[p][j])
                lowc[j]=G[p][j];
        }
    }
    return res;
}
int main(){
    int n,m;
    int x,y,w;
    while(~scanf("%d %d",&n,&m)){
        memset(G,inf,sizeof(G));
        while(m--){
            scanf("%d%d%d",&x,&y,&w);
            G[x][y]=G[y][x]=w;
        }
        printf("%d\n",prim(G,n));
    }
}

 

Kruskal

#include<iostream>    
#include<cstring>    
#include<string>    
#include<cstdio>    
#include<algorithm>    
using namespace std;    
#define MAX 50005    
int father[MAX], son[MAX];    
int v, l;    
    
typedef struct Kruskal //存储边的信息    
{    
    int a;    
    int b;    
    int value;    
};    
    
bool cmp(const Kruskal & a, const Kruskal & b)    
{    
    return a.value < b.value;    
}    
    
int unionsearch(int x) //查找根结点+路径压缩    
{    
    return x == father[x] ? x : unionsearch(father[x]);    
}    
    
bool join(int x, int y) //合并    
{    
    int root1, root2;    
    root1 = unionsearch(x);    
    root2 = unionsearch(y);    
    if(root1 == root2) //为环    
        return false;    
    else if(son[root1] >= son[root2])    
        {    
            father[root2] = root1;    
            son[root1] += son[root2];    
        }    
        else    
        {    
            father[root1] = root2;    
            son[root2] += son[root1];    
        }    
    return true;    
}    
    
int main()    
{    
    int ncase, ltotal, sum, flag;    
    Kruskal edge[MAX];    
        scanf("%d%d", &v, &l);    
        ltotal = 0, sum = 0, flag = 0;    
        for(int i = 1; i <= v; ++i) //初始化    
        {    
            father[i] = i;    
            son[i] = 1;    
        }    
        for(int i = 1; i <= l ; ++i)    
        {    
            scanf("%d%d%d", &edge[i].a, &edge[i].b, &edge[i].value);    
        }    
        sort(edge + 1, edge + 1 + l, cmp); //按权值由小到大排序    
        for(int i = 1; i <= l; ++i)    
        {    
            if(join(edge[i].a, edge[i].b))    
            {    
                ltotal++; //边数加1    
                sum += edge[i].value; //记录权值之和    
                //cout<<edge[i].a<<"->"<<edge[i].b<<endl;    
            }    
            if(ltotal == v - 1) //最小生成树条件：边数=顶点数-1    
            {    
                flag = 1;    
                break;    
            }    
        }    
        if(flag) printf("%d\n", sum);    
        else printf("data error.\n");      
    return 0;    
}