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poj1861 最小生成树 prim & kruskal

时间:2017-05-10 16:29:10      阅读:226      评论:0      收藏:0      [点我收藏+]

标签:using   can   inpu   return   class   tor   ++   content   bre   

//    poj1861 最小生成树 prim & kruskal
//
//    一个水题,为的仅仅是回味一下模板。日后好有个照顾不是



#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <iostream>

using namespace std;

const int MAX_N = 1008;
const int INF = 0x3f3f3f3f;
int g[MAX_N][MAX_N];
int n,m;
int cnt;
int mx;
int d[MAX_N];
bool vis[MAX_N];
int pre[MAX_N];
struct edge{
    int from;
    int to;
    int w;

    edge(){

    }

    edge(int from,int to,int w): from(from),to(to),w(w){

    }

};
edge edges[MAX_N * 15];

bool cmp(edge a,edge b){
    return a.w < b.w;
}

void print1(){
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++)
            printf("%d ",g[i][j]);
        puts("");
    }

}

void input1(){
    int u,v,cost;

    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++){
            g[i][j] = INF;
            if (i == j)
                g[i][j] = 0;
        }



    for (int i=0;i<m;i++){
        scanf("%d%d%d",&u,&v,&cost);
        g[u][v] = g[v][u] = cost;

    }

//    print1();

    for (int i=1;i<=n;i++){
        pre[i] = 1;
    }

    cnt = 0;

}

int fa[MAX_N];
int height[MAX_N];
void input2(){
    int u,v,cost;
    for (int i=1;i<=m;i++){
        scanf("%d%d%d",&u,&v,&cost);
        edges[i] = edge(u,v,cost);
    }
    for (int i=1;i<=n;i++){
        fa[i] = i;
        height[i] = 0;
    }
    sort(edges+1,edges+m+1,cmp);

}

int getf(int x){
    if (x==fa[x])
        return x;
    return fa[x] = getf(fa[x]);
}



void kruskal(){
    int cnt = 0;
    int mx = 0;
    for (int i=1;i<=m;i++){
        int x = getf(edges[i].from);
        int y = getf(edges[i].to);
        if (x==y)
            continue;
        if (height[x] < height[y]){
            fa[x] = y;
        }else {
            fa[y] = x;
            if (height[x]==height[y])
                height[x]++;
        }
        mx = max(mx,edges[i].w);
        edges[cnt++] = edges[i];
    }
    printf("%d\n",mx);
    printf("%d\n",cnt);
    for (int i=0;i<cnt;i++){
        printf("%d %d\n",edges[i].from,edges[i].to);
    }
}


void prim(){
    for (int i=1;i<=n;i++)
        d[i] = g[1][i];
    d[1] = 0;
    for (int i=1;i<=n;i++){
        vis[i] = 0;
    }
    vis[1] = 1;
    mx = 0;
    for (int i=1;i<=n;i++){
        int k = -1;
        for (int j=1;j<=n;j++){
            if (!vis[j] && (k == -1 || d[k] > d[j])){
                k = j;
            }
        }
        if (k==-1)
            break;
        vis[k]++;
        mx = max(mx,d[k]);
        edges[cnt++] = edge(k,pre[k],d[k]);
        for (int j=1;j<=n;j++){
            if (!vis[j] && g[k][j]!= INF &&d[j] > g[k][j]){
                d[j] = g[k][j];
                pre[j] = k;
            }
        }

    }

}

void print(){
    printf("%d\n",mx);
    printf("%d\n",cnt);
    for (int i=0;i<cnt;i++){
        printf("%d %d\n",edges[i].from,edges[i].to);
    }
}

void solve(){
    //prim();
    kruskal();
    //print();
}

int main(){

    freopen("1.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF){

        //input1();
        input2();
        solve();
    }
}


poj1861 最小生成树 prim &amp; kruskal

标签:using   can   inpu   return   class   tor   ++   content   bre   

原文地址:http://www.cnblogs.com/lxjshuju/p/6836422.html

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