杭电 1162 kruskal（）算法最小生成树

最新推荐文章于 2020-04-29 13:20:05 发布

最新推荐文章于 2020-04-29 13:20:05 发布 · 144 阅读

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#java

本文介绍了一个有趣的编程问题：通过构建最小生成树来计算连接所有绘画坐标点所需的最短线段总长度。文章提供了完整的AC代码实现，并详细解释了输入输出格式。

一道稍微有点难度的最小生成树的题，，，仔细想想的话，还是很容易做出来的。。。题目：

Eddy's picture

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2835Accepted Submission(s): 1377

Problem Description

Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture.Eddy feels very puzzled,in order to change all friends 's view to his technical of painting pictures ,so Eddy creates a problem for the his friends of you.
Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. How many distants does your duty discover the shortest length which the ink draws?

Input

The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point.

Input contains multiple test cases. Process to the end of file.

Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.

Sample Input

Sample Output

3.41

ac代码：

#include <iostream> #include <algorithm> #include <cstdio> #include <cmath> using namespace std; const int N=10005; double leftt[N],rightt[N]; //double value[N]; int father[N],r[N]; struct point{ int x,y; double value; }aa[10005]; int cmp(const int i,const int j){ return aa[i].value<aa[j].value; } int find(int x){ if(father[x]!=x) father[x]=find(father[x]); return father[x]; } int main(){ int n; while(scanf("%d",&n)!=EOF){ for(int i=0;i<N;++i){ father[i]=i; r[i]=i; aa[i].value=0.0; } /*for(int i=0;i<10;++i) {printf("%d ",r[i]);}*/ for(int i=0;i<n;++i){ scanf("%lf%lf",&leftt[i],&rightt[i]); } int k=0; for(int i=0;i<n;++i){ for(int j=0;j<n;++j){ if(i!=j) {aa[k].value=sqrt((leftt[i]-leftt[j])*(leftt[i]-leftt[j])+(rightt[i]-rightt[j])*(rightt[i]-rightt[j])); aa[k].x=i; aa[k].y=j; k++; } } } //printf("k===%d\n",k); /*for(int i=0;i<k;++i) {printf("%d ",r[i]);}*/ sort(r,r+k,cmp); /*for(int i=0;i<k;++i) {printf("%d ",r[i]);} printf("\n");*/ double sum=0; for(int i=0;i<k;++i){ int e=r[i]; int x=find(aa[e].x); int y=find(aa[e].y); if(x!=y){ sum+=aa[e].value; /*printf("x==%d\n y==%d\n",x,y); printf("sum===%.2lf\n",sum);*/ father[x]=y; } } printf("%.2lf\n",sum); } return 0; }