Highways
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 11071 | Accepted: 3145 | 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
题意：给出n个城市的坐标，再给出m，后面有m行，每行两个数表示已经有公路连接的两个城市的编号。现在要使这n个城市实现畅通，且修路的成本最低。输出全部还需要修建公路的城市，每行两个城市编码，表示这两个城市间需要修一条公路。
最小生成树，输出连通的点的组合就行了。用prim算法比较方便，kruskal需要优化。
prim算法解题，用数组记录lowcost数组中值对应的城市。
代码如下：
<span style="font-size:12px;">#include<cstdio>
#include<cstring>
#define INF 0x3f3f3f
int map[800][800],n;
void prim()
{
int i,j,next,min;
int lowcost[800],visit[800],pos[800];
memset(visit,0,sizeof(visit));
for(i=1;i<=n;++i)
{
lowcost[i]=map[1][i];
pos[i]=1;//用memset对数组初始化为1，居然结果不对，用循环就可以了，无语
}
visit[1]=1;
for(i=2;i<=n;++i)
{
min=INF;
for(j=1;j<=n;++j)
{
if(!visit[j]&&min>lowcost[j])
{
min=lowcost[j];
next=j;
}
}
if(min!=INF&&min!=0)//当前能够找到最小的权值，且不是已经连接的路径
printf("%d %d\n",pos[next],next);
visit[next]=1;
for(j=1;j<=n;++j)
{
if(!visit[j]&&lowcost[j]>map[next][j])
{
lowcost[j]=map[next][j];
pos[j]=next;//pos数组中下表j表示j城市，对应的值next表示next城市，lowcost[j]表示这两座城市的距离
}
}
}
}
int main()
{
int i,j,x[800],y[800],m,d,a,b;
while(scanf("%d",&n)!=EOF)
{
memset(map,INF,sizeof(map));
for(i=1;i<=n;++i)
scanf("%d%d",&x[i],&y[i]);
for(i=1;i<n;++i)
{
for(j=1+i;j<=n;++j)
{
d=(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
map[i][j]=map[j][i]=d;
}
}
scanf("%d",&m);
for(i=0;i<m;++i)
{
scanf("%d%d",&a,&b);
map[a][b]=map[b][a]=0;
}
prim();
}
return 0;
}
</span>
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