HDOJ 1751 Highways(最小產生樹prim)

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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

91 50 0 3 24 55 10 45 21 25 331 39 71 2

 

Sample Output

1 63 74 95 78 3

 

題意：給出n個城市的座標，再給出m，後面有m行，每行兩個數表示已經有公路串連的兩個城市的編號。現在要使這n個城市實現暢通，且修路的成本最低。輸出全部還需要修建公路的城市，每行兩個城市編碼，表示這兩個城市間需要修一條公路。

 

最小產生樹，輸出連通的點的組合就行了。用prim演算法比較方便，kruskal需要最佳化。

prim演算法解題，用數組記錄lowcost數組中值對應的城市。

 

代碼如下：

 

<span style="font-size:12px;">#include<cstdio>#include<cstring>#define INF 0x3f3f3fint 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對數組初始化為1，居然結果不對，用迴圈就可以了，無語 }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)//當前能夠找到最小的權值，且不是已經串連的路徑 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數組中下表j表示j城市，對應的值next表示next城市，lowcost[j]表示這兩座城市的距離 }}}}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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HDOJ 1751 Highways(最小產生樹prim)