hdu 4463 Outlets（图论：最小生成树）

来源：互联网发布：oracle数据库下载安装编辑：程序博客网时间：2024/06/04 20:09

典型的最小生成树，直接套prim写的

题意要求是有一条边一定要加入MST中

所以我们可以先令这条边的权值为0

最后计算出来的结果再加上这条边原来的权值

代码如下：

#include <cmath>#include <cstdio>#include <iostream>#include <algorithm>#define MAXN 1010#define LL long long#define INF 0x7fffffffusing namespace std;int n;struct POS{    double x, y;}pos[MAXN];double dis[MAXN];double g[MAXN][MAXN];double prim() {      int now, i, j;      int minnode;    double ans, minedge;      for(i=1; i<=n; ++i)          dis[i] = INF;      now = 1;      ans = 0;      for(i=1; i<n; ++i) {          dis[now] = -1;          minedge = INF;          for(j=1; j<=n; ++j) {              if(now!=j && dis[j]>=0) {                  dis[j] = min(dis[j], g[now][j]);                  if(dis[j] < minedge) {                      minedge = dis[j];                      minnode = j;                  }              }          }          now = minnode;          ans += minedge;      }      return ans;  }  double get_dis(int i, int j) {    double tmp1 = (pos[i].x-pos[j].x)*(pos[i].x-pos[j].x);    double tmp2 = (pos[i].y-pos[j].y)*(pos[i].y-pos[j].y);    return sqrt(tmp1+tmp2);}int main(void) {    int p, q;    while(scanf("%d", &n) && n) {        scanf("%d%d", &p, &q);        for(int i=1; i<=n; ++i) {            scanf("%lf%lf", &pos[i].x, &pos[i].y);        }        for(int i=1; i<=n; ++i) {            for(int j=i+1; j<=n; ++j) {                g[i][j] = g[j][i] = get_dis(i, j);            }        }        double tmp = g[p][q];        g[p][q] = g[q][p] = 0.0;        printf("%.2lf\n", prim()+tmp);    }    return 0;}

0 0