zoj 1203 Swordfish

来源：互联网发布：windows xp是什么意思编辑：程序博客网时间：2024/05/21 10:14

题意要求令总的代价最小，很明显是最小生成树。

在这里，很明显用Prim算法要比用Kruskal算法更方便

#include <iostream>#include <cstdio>#include <cmath>using namespace std;#define MAXN 108//计算出来的两城市之间距离最大值#define MAX_DOUBLE 28285struct Point{double x,y;};//计算两city之间距离double computerDist(Point A,Point B){return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));}//记录各个点之间的距离double dists[MAXN][MAXN];//记录每个端点与0号端点的最小距离double distzero[MAXN];Point cites[MAXN];//记录城市是否在最小生成树中bool used[MAXN];int n;int ct = 0;void init(){    //计算好各city之间的距离    for (int i=0;i<n;i++){        scanf("%lf %lf",&cites[i].x,&cites[i].y);        for (int j=0 ;j<i;j++){            dists[i][j] = computerDist(cites[i],cites[j]);            dists[j][i] = dists[i][j];        }        used[i] = 0;        distzero[i] = MAX_DOUBLE;    }}void prim(){    distzero[0] = true;    used[0] = true;    int now = 0;    double sum = 0;    for (int i=0;i<n-1;i++){        double min = MAX_DOUBLE;        int next = 0;        for (int j=0;j<n;j++){            if (!used[j]){                if(distzero[j]>dists[now][j]){                    distzero[j] = dists[now][j];                }                if(min>distzero[j]){                    min = distzero[j];                    next = j;                }            }        }        sum += min;        now = next;        used[now] = true;    }    printf("Case #%d:\n", ct);    printf("The minimal distance is: ");    printf("%.2lf\n", sum);}int main(){while(scanf("%d",&n)&&n!=0){ct++;//格式控制if (ct!=1) printf("\n");init();        prim();}    return 0;}