soj1169: Networking_最小生成树

soj1169: Networking

http://acm.scu.edu.cn/soj/problem.action?id=1169

简介题意：求最小生成树的权值之和。prim和kruskal都可以，我都写了，不过kruskal数组开小了re了好几次，简直太愚蠢，大家要注意看题，本道题非常简单。

首先prim：

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define MAX 55#define INF 0x3f3f3f3fint map[MAX][MAX];int dist[MAX];bool visit[MAX];void prim(int n){    memset(dist,INF,sizeof(dist));    memset(visit,0,sizeof(visit));    for(int i = 1;i <= n; i++)        if(map[1][i] != 0)            dist[i] = map[1][i];    visit[1] = 1;    int price = 0;    for(int i = 1;i <= n; i++)    {        int u;        int min = INF;        for(int j = 1;j <= n; j++)            if(!visit[j] && dist[j]<min )            {                min = dist[j];                u = j;            }        if(min < INF)        {            visit[u] = 1;            price += min;            for(int j = 1;j <= n; j++)                if(!visit[j] && map[u][j]!=0 && dist[j]>map[u][j])                    dist[j] = map[u][j];        }        else break;    }    printf("%d\n",price);}int main(){    int n,m;    while(~scanf("%d",&n)&&n)    {        scanf("%d",&m);        memset(map,0,sizeof(map));        for(int i = 1;i <= m; i++)        {            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            if(map[a][b] == 0)                map[a][b] = map[b][a] = c;            else if(map[a][b] > c)                map[a][b] = map[b][a] = c;        }        prim(n);    }    return 0;}

没什么好解释的，不过又排了一次第一，高兴~

然后接下来kruskal：

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define MAX 110#define INF 0x3f3f3f3fint pre[MAX];int map[MAX][MAX];int visit[MAX][MAX];struct Node{    int x,y,len;}edge[MAX];bool cmp(Node a, Node b){    return a.len < b.len;}int find(int a){    if(pre[a] == a)        return a;    else        return pre[a] = find(pre[a]);}void Kruskal(int n,int k){    for(int i = 0;i <= n; i++)        pre[i] = i;    int price = 0;    for(int i = 1;i <= k; i++)    {        int a = find(edge[i].x);        int b = find(edge[i].y);        if(a != b)        {            pre[a] = b;            price += edge[i].len;        }    }    printf("%d\n",price);}int main(){    int n,m;    while(~scanf("%d",&n)&&n)    {        scanf("%d",&m);        memset(map,0,sizeof(map));        memset(visit,0,sizeof(visit));        for(int i = 1;i <=m ; i++)        {            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            if(map[a][b] == 0)                map[a][b] = map[b][a] = c;            else if(map[a][b] > c)                map[a][b] = map[b][a] = c;        }        int k = 1;        for(int i = 1;i <= n; i++)        for(int j = 1;j <= n; j++)        {            if(!visit[i][j] && map[i][j] != 0)            {                edge[k].x = i;                edge[k].y = j;                edge[k++].len = map[i][j];                visit[i][j] = visit[j][i] = 1;            }        }        sort(edge+1,edge+k,cmp);        Kruskal(n,k-1);    }    return 0;}

kruskal主要看数组啊！！！我刚才没有改MAX，所以re好多次，因为kruskal写的时候要注意本题可以两点之间很多条路线，所以记录最小的即可。要注意哦。

判断用哪一种方法，你就看点多不多，多的话你就用kruskal，少的话就prim