uva 10397 - Connect the Campus

Problem E

Connect the Campus

Input: standard input

Output: standard output

Time Limit: 2 seconds

Many new buildings are under construction on the
campus of the University of Waterloo.
The university has hired bricklayers, electricians, plumbers, and a computer
programmer. A computer programmer? Yes, you have been hired to ensure that each
building is connected to every other building (directly or indirectly) through
the campus network of communication cables.

We will treat each building as a point specified
by an x-coordinate and a y-coordinate. Each communication cable connects
exactly two buildings, following a straight line between the buildings.
Information travels along a cable in both directions. Cables can freely cross
each other, but they are only connected together at their endpoints (at
buildings).

You have been given a campus map which shows the
locations of all buildings and existing communication cables. You must not
alter the existing cables. Determine where to install new communication cables
so that all buildings are connected. Of course, the university wants you to
minimize the amount of new cable that you use.

Fig: University
of Waterloo Campus

Input

The input file describes several test case. The description of each test case is given
below:

The first line of each test case contains the
number of buildings N (1<=N<=750).
The buildings are labeled from 1 to N. The next N lines give the x and y coordinates of the buildings. These
coordinates are integers with absolute values at most 10000. No two buildings occupy the same point. After that there is
a line containing the number of existing cables M (0 <= M <= 1000) followed by M lines describing the existing cables. Each cable is represented
by two integers: the building numbers which are directly connected by the
cable. There is at most one cable directly connecting each pair of buildings.

Output

For each set of input, output in a single line
the total length of the new cables that you plan to use, rounded to two decimal
places.

Sample Input

4

103 104

104 100

104 103

100 100

1

4 2

4

103 104

104
100

104
103

100
100

1

4 2

Sample Output4.41

4.41

#include <iostream>
#include <stack>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <fstream>
#include <stack>
#include <list>
#include <sstream>
#include <cmath>

using namespace std;

#define ms(arr, val) memset(arr, val, sizeof(arr))
#define mc(dest, src) memcpy(dest, src, sizeof(src))
#define N 755
#define INF 0x3fffffff
#define vint vector<int>
#define setint set<int>
#define mint map<int, int>
#define lint list<int>
#define sch stack<char>
#define qch queue<char>
#define sint stack<int>
#define qint queue<int>
/*最小生成树算法prim:稠密图(n*n)，kruskal:稀疏图(e*lge)*/

struct point
{
int x, y;
}in
;

bool visit
;
double low
;
double g

;
int n, m;

int multi2(int x1, int x2, int y1, int y2)
{
int x = x1 - x2;
int y = y1 - y2;
return x * x + y * y;
}

void prim()
{
ms(visit, 0);
double ans = 0;
int s = n - 1, _minp;
double _min;
for (int i = 1; i <= n; i++)
{
low[i] = g[1][i];
}
low[1] = INF;
visit[1] = 1;
while (s--)
{
_min = low[1];
for (int i = 1; i <= n; i++)//寻找最小边
{
if (!visit[i] && low[i] < _min)
{
_min = low[i];
_minp = i;
}
}
visit[_minp] = true;
ans += sqrt(_min);
for (int i = 1; i <= n; i++)//更新low数组
{
if (!visit[i] && g[i][_minp] < low[i])
{
low[i] = g[i][_minp];
}
}
}
printf("%.2lf\n", ans);
}

int main()
{
while (~scanf("%d", &n))
{
for (int i = 1; i <= n; i++)
{
scanf("%d%d", &in[i].x, &in[i].y);
for (int j = 1; j < i; j++)
{
g[j][i] = g[i][j] = multi2(in[i].x, in[j].x, in[i].y, in[j].y);
}
}
ms(visit, 0);
int s, e;
scanf("%d", &m);
while (m--)
{
scanf("%d%d", &s, &e);
g[s][e] = g[e][s] = 0;//将已经连通的点之间的距离变为0，在prim算法中最短边肯定会被选择(关键之处)
}
prim();
}
return 0;
}