Highways
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 6078 Accepted: 1650 Special Judge
Description
The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has a very poor system of public highways. The Flatopian government is aware of this problem and has already constructed a number of highways connecting some of the most important towns. However, there are still some towns that you can’t reach via a highway. It is necessary to build more highways so that it will be possible to drive between any pair of towns without leaving the highway system.
Flatopian towns are numbered from 1 to N and town i has a position given by the Cartesian coordinates (xi, yi). Each highway connects exaclty two towns. All highways (both the original ones and the ones that are to be built) follow straight lines, and thus their length is equal to Cartesian distance between towns. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways.
The Flatopian government wants to minimize the cost of building new highways. However, they want to guarantee that every town is highway-reachable from every other town. Since Flatopia is so flat, the cost of a highway is always proportional to its length. Thus, the least expensive highway system will be the one that minimizes the total highways length.
Input
The input consists of two parts. The first part describes all towns in the country, and the second part describes all of the highways that have already been built.
The first line of the input file contains a single integer N (1 <= N <= 750), representing the number of towns. The next N lines each contain two integers, xi and yi separated by a space. These values give the coordinates of ith town (for i from 1 to N). Coordinates will have an absolute value no greater than 10000. Every town has a unique location.
The next line contains a single integer M (0 <= M <= 1000), representing the number of existing highways. The next M lines each contain a pair of integers separated by a space. These two integers give a pair of town numbers which are already connected by a highway. Each pair of towns is connected by at most one highway.
Output
Write to the output a single line for each new highway that should be built in order to connect all towns with minimal possible total length of new highways. Each highway should be presented by printing town numbers that this highway connects, separated by a space.
If no new highways need to be built (all towns are already connected), then the output file should be created but it should be empty.
Sample Input
9
1 5
0 0
3 2
4 5
5 1
0 4
5 2
1 2
5 3
3
1 3
9 7
1 2
Sample Output
1 6
3 7
4 9
5 7
8 3
解题心得：
- 题目很简单，就是跑一个最小生成树，然后记录需要建立的新边。但是很坑啊，题目中说如果没有输出那么就会建立一个空白的文件，所以如果是写的多组输入，就会WA，不知道为啥，可能是没有建立空白的新文件吧。
- 然后就是邪最小生成树，两种写法
- Kruskal算法先得出每一条边，然后对边排序，从小的边开始选择，用并查集来判断是否形成了环。
- Prim算法是选择一个点，然后找出距离这个点最近的一个点，连成边，然后以找出的店为目标，再从没被连接的点中找出一个距离最近的点，连成边，然后一直将所有的点全部连接。
Kruskal算法代码：
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
const int maxn = 1e3+100;
struct node
{
int x,y;
} p[maxn*maxn];
struct Path
{
int s,e,len;
} path[maxn*maxn];
int n,m,father[maxn*maxn],ans,t;
vector <pair<int,int> > ve;
bool cmp(Path a,Path b)
{
return a.len<b.len;
}
int find(int x)
{
if(father[x] == x)
return x;
return father[x] = find(father[x]);
}
void merge(int x,int y)
{
int fx = find(x);
int fy = find(y);
if(fx != fy)
father[fy] = fx;
}
int dis(int x,int y)
{
int d = (p[x].x - p[y].x)*(p[x].x - p[y].x) + (p[x].y - p[y].y)*(p[x].y - p[y].y);
return d;
}
void init()
{
ans = t = 0;
for(int i=1; i<=n; i++)
{
father[i] = i;
scanf("%d%d",&p[i].x,&p[i].y);
}
//枚举每一条边
for(int i=1; i<=n; i++)
for(int j=i+1; j<=n; j++)
{
path[t].s = i;
path[t].e = j;
path[t++].len = dis(i,j);
}
sort(path,path+t,cmp);//将边从小到大排序
cin>>m;
for(int i=0; i<m; i++)
{
int a,b;
scanf("%d%d",&a,&b);
if(find(a) != find(b))
merge(a,b);//将已经有路的点合并
}
}
void solve()
{
for(int i=0; i<t; i++)
{
int x = path[i].s;
int y = path[i].e;
int len = path[i].len;
if(find(x) != find(y))//如果不是同一个祖先那么连接就不会形成环
{
ans += len;
merge(x,y);
ve.push_back(make_pair(x,y));//记录需要连接的点
}
}
for(int i=0; i<ve.size(); i++)
{
pair<int,int> p;
p = ve[i];
printf("%d %d
",p.first,p.second);
}
ve.clear();
}
int main()
{
cin>>n;
init();
solve();
return 0;
}Prim算法代码
#include<stdio.h>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 1000;
//lowcast记录的是各点距离已经生成了的树的距离
int maps[maxn][maxn],lowcost[maxn],n,m,Edge[maxn];
bool vis[maxn];//记录点是否已经在树中
struct NODE
{
int x,y;
}node[maxn];
int get_dis(int x,int y)
{
int dis = (node[x].x - node[y].x)*(node[x].x - node[y].x) + (node[x].y - node[y].y)*(node[x].y - node[y].y);
return dis;
}
void init()
{
cin>>n;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&node[i].x,&node[i].y);
for(int j=1;j<i;j++)
maps[i][j] = maps[j][i] = get_dis(i,j);//记录两点之间的距离
maps[i][i] = 0x3f3f3f3f;//不可能自身到自身
}
memset(vis,0,sizeof(vis));//记录该点是否已经在树上
vis[1] = 1;
cin>>m;
while(m--)
{
int a,b;
scanf("%d%d",&a,&b);
maps[a][b] = maps[b][a] = 0;
}
for(int i=1;i<=n;i++)
{
lowcost[i] = maps[i][1];//先得到所有点距离第一个点的距离
Edge[i] = 1;
}
}
void Prim()
{
for(int i=1;i<n;i++)
{
int Min = 0x3f3f3f3f;
int point;
for(int j=1;j<=n;j++)//当前树距离最近的点
if(!vis[j] && Min > lowcost[j])
{
Min = lowcost[j];
point = j;
}
vis[point] = true;//将这个点加入树中
for(int k=1;k<=n;k++)
{
if(!vis[k] && lowcost[k] > maps[point][k])
{
Edge[k] = point;//记录添加边的两个点
lowcost[k] = maps[point][k];
}
}
if(maps[Edge[point]][point])
printf("%d %d
",Edge[point],point);
}
}
int main()
{
init();
Prim();
return 0;
}