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Highways
Time Limit: 1000MS | Memory Limit: 10000K | |||
|---|---|---|---|---|
Total Submissions: 14613 | Accepted: 4211 | 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 i th 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
代码语言：javascript复制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 2Sample Output
代码语言：javascript复制1 6
3 7
4 9
5 7
8 3code:
prim算法
代码语言：javascript复制#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<sstream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
const int maxn=755;
const int INF=1000000;
double graph[maxn][maxn];
double lowcost[maxn];/*lowcost表示每个点的最小花费;*/
int closet[maxn];/*closet表示最小花费对应相连的点*/
int visited[maxn];/*visited区分两个集合*/
int n;/*n个点*/
struct dot{
double x,y;
}a[maxn];
double f(dot p,dot q){
return sqrt((p.x-q.x)*(p.x-q.x) (p.y-q.y)*(p.y-q.y));
}
void createGraph(){
memset(graph,0,sizeof(graph));
memset(lowcost,0,sizeof(lowcost));
memset(closet,0,sizeof(closet));
memset(visited,0,sizeof(visited));
for(int i=0;i<n;i )
for(int j=0;j<n;j ){
if(i==j) graph[i][j]=INF;
else graph[i][j]=graph[j][i]=f(a[i],a[j]);
}
}
void prim(){
visited[0]=1;/*选中第一个点*/
for(int i=0;i<n;i ){
lowcost[i]=graph[i][0];/*每个点与第一个点的权值*/
closet[i]=0;/*与i点相连的是第一个点*/
}
for(int i=1;i<n;i ){ /*剩下n-1个点*/
int k=0;
double minn=lowcost[0];
for(int j=0;j<n;j ){
if(!visited[j] && lowcost[j]<minn){
minn=lowcost[j];
k=j;
}
}
if(graph[k][closet[k]]!=0) printf("%d %dn",k 1,closet[k] 1);
visited[k]=1;
for(int t=0;t<n;t ){ /*松弛操作*/
if(!visited[t] && lowcost[t]>graph[t][k]){
lowcost[t]=graph[t][k];
closet[t]=k;
}
}
}
}
int main()
{
// freopen("input.txt","r",stdin);
int m;
scanf("%d",&n);
for(int i=0;i<n;i ){
scanf("%lf%lf",&a[i].x,&a[i].y);
}
createGraph();
scanf("%d",&m);
int b,c;
while(m--){
scanf("%d%d",&b,&c);
graph[b-1][c-1]=graph[c-1][b-1]=0;
}
prim();
}Kruskal算法：注意要用G 提交
代码语言：javascript复制#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<sstream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
const int MAXN = 755; /*结点数目上限*/
int pa[MAXN]; /*pa[x]表示x的父节点*/
int rank[MAXN]; /*rank[x]是x的高度的一个上界*/
int flag;
struct node{
int x,y;
double w;
}edge[MAXN*MAXN];
struct dot{
double x,y;
}a[MAXN];
bool cmp(node p,node q){
return p.w<q.w;
}
double f(dot p,dot q){
return sqrt((p.x-q.x)*(p.x-q.x) (p.y-q.y)*(p.y-q.y));
}
/*创建一个单元集*/
void make_set(int x)
{
pa[x] = x;
rank[x] = 0;
}
/*带路径压缩的查找*/
int find_set(int x)
{
if(x != pa[x])
pa[x] = find_set(pa[x]);
return pa[x];
}
/*按秩合并x，y所在的集合*/
void union_set(int xx, int yy,double w)
{
int x = find_set(xx);
int y = find_set(yy);
if(x == y)return ;
if(rank[x] > rank[y])/*让rank比较高的作为父结点*/
{
pa[y] = x;
}
else
{
pa[x] = y;
if(rank[x] == rank[y])
rank[y] ;
}
if(w!=0){
printf("%d %dn",xx,yy);
flag=1;
}
}
int main()
{
// freopen("input.txt","r",stdin);
int n,m;
scanf("%d",&n);
for(int i=0;i<MAXN;i )
make_set(i);
for(int i=1;i<=n;i ){
scanf("%lf%lf",&a[i].x,&a[i].y);
}
int k=0;
for(int i=1;i<=n;i ){
for(int j=i 1;j<=n;j ){
edge[k].x=i;
edge[k].y=j;
edge[k].w=f(a[i],a[j]);
k ;
}
}
scanf("%d",&m);
int b,c;
while(m--){
scanf("%d%d",&b,&c);
union_set(b,c,0);
}
sort(edge,edge k,cmp);
flag=0;
for(int i=0;i<k;i ){
union_set(edge[i].x,edge[i].y,edge[i].w);
}
if(!flag) printf("n");
}发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/129970.html原文链接：https://javaforall.cn
