标签：foreign opp string 算法 get mos number lse pen
InputThere are several test cases. For each test case: The first line is an integer N( 3 <= N <= 50) , meaning there are N stores in the outlets. These N stores are numbered from 1 to N. The second line contains two integers p and q, indicating that the No. p store is a Nike store and the No. q store is an Apple store. Then N lines follow. The i-th line describes the position of the i-th store. The store position is represented by two integers x,y( -100<= x,y <= 100) , meaning that the coordinate of the store is (x,y). These N stores are all located at different place. The input ends by N = 0.
OutputFor each test case, print the minimum total road length. The result should be rounded to 2 digits after decimal point.
Sample Input
4
2 3
0 0
1 0
0 -1
1 -1
0
Sample Output
3.41
题意：给你n个点，最小的价值使得所有的点连通，但是p，q一定是直连的。
这是一道比较的模板的最小生成树的题，但是要保证有一条边一定在这颗树内，我们可以使用Kruskal算法的时候，直接把ans先设置为p，q之间距离的值，然后在加边的时候把值设置为0，那么根据Kruskal算法的思想，
这个边最小肯定是最先加进来了，那么其他的就和其他的没有区别了。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=105;
int n,p,q;
int cnt;
struct Point
{
int x,y;
}point[maxn];
struct Node
{
int from,to;
double value;
}node[maxn*maxn];
int fa[maxn];
bool cmp(Node a,Node b)
{
return a.value<b.value;
}
void init()
{
for(int i=0;i<maxn;i++)
fa[i]=i;
}
int findd(int x)
{
if(fa[x]==x)
return x;
else
return fa[x]=findd(fa[x]);
}
double getdist(Point a,Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double Kruskal()
{
double ans=getdist(point[p],point[q]);
for(int i=1;i<=cnt;i++)
{
int fx=findd(node[i].from);
int fy=findd(node[i].to);
if(fx!=fy)
{
ans+=node[i].value;
fa[fx]=fy;
}
}
return ans;
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
if(n==0)
break;
scanf("%d %d",&p,&q);
for(int i=1;i<=n;i++)
scanf("%d %d",&point[i].x,&point[i].y);
cnt=0;
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
cnt++;
node[cnt].from=i;node[cnt].to=j;
node[cnt].value=getdist(point[i],point[j]);
if((i==p&&j==q)||(i==q&&j==p))
node[cnt].value=0;
}
}
init();
sort(node+1,node+cnt+1,cmp);
double sum=Kruskal();
printf("%.2f\n",sum);
}
return 0;
}
标签：foreign opp string 算法 get mos number lse pen
原文地址：https://www.cnblogs.com/jkzr/p/10021561.html
