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 Nlines 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 Output
4.41
4.41


(Problem-setters: G. Kemkes & G. V. Cormack, CS Dept, University of Waterloo)

分析

    就是最小生成树,只不过要对输入进行处理:已经连接的两点合并到一起,这里用的是Kruskal算法。下面是代码:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = 800;
struct node
{
    int x;
    int y;
    int vis;
} G[maxn];
int u[maxn*maxn],v[maxn*maxn];
int r[maxn*maxn],p[maxn];
double w[maxn*maxn];
int n,m;
int cmp(const int i,const int j)
{
    return w[i] < w[j];
}
int find(int x)
{
    if(x != p[x])
    {
        p[x] = find(p[x]);
    }
    return p[x];
}
double Kruskal(int num)
{
    double ans = 0;
    sort(r+1,r+num,cmp);
    for(int i = 1; i < num; i++)
    {
        int e = r[i];
        int x = find(u[e]);
        int y = find(v[e]);
        if(x != y)
        {
            ans += w[e];
            p[x] = y;
        }
    }
    return ans;
}
int main()
{
    while(scanf("%d",&n) != EOF)
    {
        int x,y;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d",&x,&y);
            G[i].x = x;
            G[i].y = y;
            G[i].vis = 1;
            p[i] = i;
        }
        scanf("%d",&m);
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d",&x,&y);
            int px = find(x);
            int py = find(y);
            if(px != py)
            {
                p[px] = py;
            }
        }
        double dx,dy;
        int num = 1;
        for(int i = 1; i <= n; i++)
        {
            for(int j = i + 1; j <= n; j++)
            {
                u[num] = i;
                v[num] = j;
                r[num] = num;
                dx = G[i].x - G[j].x;
                dy = G[i].y - G[j].y;
                w[num++] = sqrt(dx*dx + dy*dy);
            }
        }
        printf("%.2lf\n",Kruskal(num));
    }
}



我做了一个安卓画板app现在添加了一个从url下载图片并设置成画板背板的功能。我把代码给你看,在我输入了https://www.nottingham.edu.cn/image-library/campus-news/signofnottingham.jpg的网站后提示下载失败 /* * Copyright (C) 2007 The Android Open Source Project * * Licensed under the Apache License, Version 2.0 (the "License"); * you may not use this file except in compliance with the License. * You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * See the License for the specific language governing permissions and * limitations under the License. */ package cn.edu.nottingham.jianxingwang.fingerpainter; import android.content.ContentResolver; import android.content.Context; import android.graphics.Bitmap; import android.graphics.BitmapFactory; import android.graphics.Canvas; import android.graphics.Color; import android.graphics.Paint; import android.graphics.Path; import android.net.Uri; import android.os.Bundle; import android.os.Parcelable; import android.util.AttributeSet; import android.util.Log; import android.view.MotionEvent; import android.view.View; import android.widget.Button; import java.io.File; import java.io.FileInputStream; import java.io.FileOutputStream; import java.io.IOException; import java.io.InputStream; import java.util.Stack; /** * Created by pszmdf on 13/10/16. * * Derived from android graphics API sample com.example.android.apis.graphics.Fingerpaint * android.googlesource.com/platform/development/+/master/samples/ApiDemos/src/com/example/android/apis/graphics/FingerPaint.java * */ public class FingerPainterView extends View { private Context context; private Canvas canvas; private Paint paint; private Bitmap bitmap; private Path path; private Uri uri; private Stack<Bitmap> undoStack = new
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03-29
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