HDU 1102 最小生成树裸题，kruskal，prim

2、总结：

题意：修路，裸题

(1)kruskal

//kruskal
#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
#include<cstdio>
#define max(a,b) a>b?a:b
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f
const int N=110;

int n,k;
int father[N];

struct Eage
{
    int st,en,val;
}eage[N*(N+1)/2];

bool cmp(Eage a,Eage b)
{
    return a.val<b.val;
}

int findn(int x)   //即并查集的查找
{
    int t=x;
    while(x!=father[x]){
        x=father[x];
    }
    int p;
    while(t!=x){
        p=t;
        t=father[t];
        father[p]=x;
    }
    return x;
}

int kruskal()
{
    int sum=0;
    sort(eage,eage+k,cmp);
    for(int i=0;i<k;i++){
        int x=findn(eage[i].st);
        int y=findn(eage[i].en);
        if(x!=y){
            sum+=eage[i].val;
            x=findn(x);
            y=findn(y);
            father[x]=y;
        }
    }
    return sum;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int m;
        k=0;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                scanf("%d",&m);
                if(i<j){
                    eage[k].st=i;   //st边起点，en边终点，k为边的条数
                    eage[k].en=j;
                    eage[k++].val=m;
                }
            }
        }

        for(int i=1;i<=n;i++){
            father[i]=i;
        }
        int q,a,b;
        scanf("%d",&q);
        while(q--){
            scanf("%d%d",&a,&b);
            a=findn(a);
            b=findn(b);
            father[a]=findn(b);  //把a与b连通
        }

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

    return 0;
}

View Code

(2)prim

//prim
#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
#include<cstdio>
#define max(a,b) a>b?a:b
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f
const int N=110;

int n;
int mapn[N][N];
int visit[N],dis[N];   //dis存储每个点到所选结点的最短距离

int prim()
{
    int sum=0;
    memset(visit,0,sizeof(visit));
    visit[1]=1;
    for(int i=1;i<=n;i++){  //先选第一个结点
        dis[i]=mapn[1][i];
    }
    for(int i=1;i<n;i++)    //注，<不是<=
    {
        int next,minn=INF;
        for(int j=1;j<=n;j++){  //找到下一个最近的结点
            if(!visit[j]&&minn>dis[j]){
                next=j;
                minn=dis[j];
            }
        }
        sum+=minn;
        visit[next]=1;

        for(int j=1;j<=n;j++){  //更新dis
            if(!visit[j]&&dis[j]>mapn[next][j]){
                dis[j]=mapn[next][j];
            }
        }
    }
    return sum;
}

int main()
{
    int q,a,b;
    while(scanf("%d",&n)!=EOF)
    {
        memset(mapn,INF,sizeof(mapn));
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                scanf("%d",&mapn[i][j]);
            }
        }

        scanf("%d",&q);
        while(q--){
            scanf("%d%d",&a,&b);
            mapn[a][b]=mapn[b][a]=0;
        }

        int ans=prim();
        printf("%d
",ans);
    }

    return 0;
}

View Code