ZOJ-1203

来源：互联网发布：mysql存储过程怎么用编辑：程序博客网时间：2024/06/06 15:39

早上本来就准备挑道简单的做做的，一看这题提交和AC都很多，果断搞起。前面一大堆介绍不用看，其实我也没看懂= =，题意就是给平面上一些点然后求最小生成树，直接上kruskal吧，这题的不相交集我是用数组维护的，以前都是用结构，发现数组方便很多，又优化了一点点，哈哈。做的时候函数命名为distance和STL函数名冲突，一堆编译错误搞的我莫名其妙，查了查才知道。。又坑了一回

#include<cstdio>#include<vector>#include<cmath>#include<algorithm>#include<cstring>using namespace std;namespace{    struct Edge    {        size_t u, v;        double d;    };    double dist(const pair<double, double> &p1, const pair<double, double> &p2)    {        return sqrt(                (p1.first - p2.first) * (p1.first - p2.first)                        + (p1.second - p2.second) * (p1.second - p2.second));    }    bool cmp(Edge *e1, Edge *e2)    {        return e1->d < e2->d;    }    int p[100], rank[100];    int find_set(int x)    {        if (x != p[x])            p[x] = find_set(p[x]);        return p[x];    }    void link(int x, int y)    {        if (rank[x] > rank[y])            p[y] = x;        else        {            p[x] = y;            if (rank[x] == rank[y])                rank[y]++;        }    }    void union_set(int x, int y)    {        link(find_set(x), find_set(y));    }}int main(){    int n, cs = 0;    vector<pair<double, double> > V;    vector<Edge *> E;    while (scanf("%d", &n), n)    {        if (cs)            putchar('\n');        V.clear();        E.clear();        double x, y;        for (int i = 0; i < n; i++)        {            scanf("%lf %lf", &x, &y);            V.push_back(make_pair(x, y));        }        for (size_t i = 0; i < V.size(); i++)            for (size_t j = i + 1; j < V.size(); j++)            {                Edge *e = new Edge();                e->u = i;                e->v = j;                e->d = dist(V[i], V[j]);                E.push_back(e);            }        sort(E.begin(), E.end(), cmp);        memset(rank, 0, sizeof(rank));        for (int i = 0; i < n; i++)            p[i] = i;        double res = 0;        for (size_t i = 0; i < E.size(); i++)            if (find_set(E[i]->u) != find_set(E[i]->v))            {                res += E[i]->d;                union_set(E[i]->u, E[i]->v);            }        printf("Case #%d:\n", ++cs);        printf("The minimal distance is: %.2lf\n", res);        for (size_t i = 0; i < E.size(); i++)            delete E[i];    }    return 0;}

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