POJ 1751 Highways

最小生成树的水题，直接的最小生成树算法就可以了。题意给定n个城市，只不过是又给定了已经连接起来的道路，由于输出的是城镇的编号，所以不用计算两城镇之间的真正距离（既不用算sqrt）,用的是kruskal，比较费时，估计用prim可以节省不少时间.

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <memory.h>
using namespace std;
const int MAXN = 800;
const int MAXE = 500000;
const int INF = 0x7fffffff;
int nv,ne,index,set[MAXN],pos[MAXE];
struct Edge{
    int u,v;
    int val;
}edge[MAXE];
struct Node{
    int x,y;
}node[MAXN];
int cmp(const Edge& a,const Edge& b)
{
    return a.val < b.val;
}
int find_set(int x)
{
    int root = x,tmp;
    while(set[root] >= 0)
        root = set[root];
    while( x != root)
    {
        tmp = set[x];
        set[x] = root;
        x = tmp;
    }
    return root;
}
void union_set(const int& root1,const int& root2)
{
    int a = set[root1];
    int b = set[root1];
    if( a>b)
    {
        set[root1] = root2;
        set[root2] += a;
    }else{
        set[root2] = root1;
        set[root1] += b;
    }
}
int kruskal()
{
    index = 0;
    int sum = 0,root1,root2,tmp,i;
    for(i = 0;i < ne; ++i)
    {
        root1 = find_set(edge[i].u);
        root2 = find_set(edge[i].v);
        if( root1 != root2)
        {
            pos[index] = i;
            ++index;
            union_set(root1,root2);
        }
    }
    return sum;
}
int main()
{
    int i,j,k,a,b,tmp,val,n,m,root1,root2;
    while(scanf("%d",&nv) != EOF)
    {
        memset(set,-1,sizeof(set));
        for(i = 1;i <= nv; ++i)
            scanf("%d%d",&(node[i]).x,&(node[i].y));
        index = 0;
        for(i = 1;i <= nv; ++i)
            for(j = i + 1;j <= nv; ++j)
            {
                edge[index].u = i;
                edge[index].v = j;
                edge[index].val = (node[i].x-node[j].x)*(node[i].x-node[j].x)+
                    (node[i].y - node[j].y)*(node[i].y-node[j].y);
                ++index;
            }
        ne = index;
        sort(edge,edge+ne,cmp);
        scanf("%d",&n);
        for(i = 1;i <= n; ++i)
        {
            scanf("%d%d",&a,&b);
            root1 = find_set(a);
            root2 = find_set(b);
            if( root1 != root2)
                union_set(root1,root2);
        }
        kruskal();
        for(i = 0;i < index; ++i)
            printf("%d %d\n",edge[pos[i]].u,edge[pos[i]].v);
    }
    return 0;
}