Outlets
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3165 Accepted Submission(s): 1471
Problem Description
In China, foreign brand commodities are often much more expensive than abroad. The main reason is that we Chinese people tend to think foreign things are better and we are willing to pay much for them. The typical example is, on the United Airline flight, they give you Haagendazs ice cream for free, but in China, you will pay $10 to buy just a little cup.
So when we Chinese go abroad, one of our most favorite activities is shopping in outlets. Some people buy tens of famous brand shoes and bags one time. In Las Vegas, the existing outlets can't match the demand of Chinese. So they want to build a new outlets in the desert. The new outlets consists of many stores. All stores are connected by roads. They want to minimize the total road length. The owner of the outlets just hired a data mining expert, and the expert told him that Nike store and Apple store must be directly connected by a road. Now please help him figure out how to minimize the total road length under this condition. A store can be considered as a point and a road is a line segment connecting two stores.
Input
There 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.
Output
For 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
Source
2012 Asia Hangzhou Regional Contest
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感觸良多，本來這是一道kruskal水題，但是WA*n ，交叉測試後才發現是強制轉換浮點哪裏出來問題，可能是qsort出錯，可能是其他的錯誤，AC和WA都放在下面，等qsort再練熟一點看看是怎麼回事
//AC：
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;
#define MAXN 100
#define MAXN_E 500000 //邊可能很多
struct P
{
int X, Y;
}Point[MAXN * 2];
struct A_Edge_B
{
int A, B; //A->B
int length; //Cost
}Edge[MAXN_E];
int k;
int Par[MAXN], Rank[MAXN];
void Init(int n)
{
for (int i = 1;i <= n;i++)
{
Par[i] = i;
Rank[i] = 0;
}
}
int Find(int x)
{
if (Par[x] == x) return x;
else return Par[x] = Find(Par[x]);
}
void Unite(int x, int y)
{
x = Find(x);y = Find(y);
if (x == y) return;
if (Rank[x]<Rank[y]) Par[x] = y;
else Par[y] = x;
if (Rank[x] == Rank[y]) Rank[x]++;
}
int cmp(const void *a, const void *b)
{
A_Edge_B *c, *d;
c = (A_Edge_B *)a;
d = (A_Edge_B *)b;
return c->length - d->length;
}
double Cal_Len(int a,int b)//a->b
{
int dX = Point[a].X - Point[b].X;
int dY = Point[a].Y - Point[b].Y;
return dX * dX + dY * dY;
}
double Handle(int n)
{
int Nike, Apple;
double ans;
Init(n);
scanf("%d%d", &Nike, &Apple);
for (int i = 1;i<=n;i++) scanf("%d%d", &Point[i].X, &Point[i].Y);
k = 0;
for (int i = 1;i<=n;i++)
for (int j = 1;j<=i;j++)
{
k++;
Edge[k].A = i;
Edge[k].B = j;
Edge[k].length = Cal_Len(i,j);
}
qsort(Edge+1, k, sizeof(Edge[0]),cmp);
ans = sqrt((double)Cal_Len(Nike,Apple));
Unite(Nike, Apple);
for (int i = 1;i<=k;i++)
{
int Pa = Find(Edge[i].A);
int Pb = Find(Edge[i].B);
if (Pa == Pb) continue;
Unite(Pa, Pb);
ans += sqrt((double)Edge[i].length);
}
return ans;
}
int main()
{
// freopen("F:\\test.txt", "r", stdin);
int n;
while (~scanf("%d", &n) && n)
printf("%.2lf\n", Handle(n));
return 0;
}
//WA
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;
#define MAXN 100
#define MAXN_E 500000 //邊可能很多
int Par[MAXN], Rank[MAXN];
struct P
{
int X, Y;
}Point[MAXN * 2];
struct A_Edge_B
{
int A, B; //A->B
double length; //Cost
}Edge[MAXN_E];
int k;
void Init(int n)
{
for (int i = 1;i <= n;i++)
{
Par[i] = i;
Rank[i] = 0;
}
}
int Find(int x)
{
if (Par[x] == x) return x;
else return Par[x] = Find(Par[x]);
}
void Unite(int x, int y)
{
x = Find(x);y = Find(y);
if (x == y) return;
if (Rank[x]<Rank[y]) Par[x] = y;
else Par[y] = x;
if (Rank[x] == Rank[y]) Rank[x]++;
}
int cmp(const void *a, const void *b)
{
A_Edge_B *c, *d;
c = (A_Edge_B *)a;
d = (A_Edge_B *)b;
return c->length - d->length >=0? 1:0;
}
double Cal_Len(int a,int b)//a->b
{
int dX = Point[a].X - Point[b].X;
int dY = Point[a].Y - Point[b].Y;
return sqrt(double(dX * dX + dY * dY));
}
double Handle(int n)
{
int Nike, Apple;
double ans;
Init(n);
scanf("%d%d", &Nike, &Apple);
for (int i = 1;i<=n;i++) scanf("%d%d", &Point[i].X, &Point[i].Y);
k = 0;
for (int i = 1;i<=n;i++)
for (int j = 1;j<=i;j++)
{
k++;
Edge[k].A = i;
Edge[k].B = j;
Edge[k].length = Cal_Len(i,j);
}
qsort(Edge+1, k, sizeof(Edge[0]),cmp);
ans = Cal_Len(Nike,Apple);
Unite(Nike, Apple);
for (int i = 1;i<=k;i++)
{
int Pa = Find(Edge[i].A);
int Pb = Find(Edge[i].B);
if (Pa == Pb) continue;
Unite(Pa, Pb);
ans += Edge[i].length;
}
return ans;
}
int main()
{
//freopen("F:\\test.txt", "r", stdin);
int n;
while (~scanf("%d", &n) && n)
printf("%.2lf\n", Handle(n));
return 0;
}