POJ1251 Jungle Roads Kruskal+scanf输入小技巧
Jungle Roads
The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems.
Input
Output
Sample Input
9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18 G 55
D 1 E 44
E 2 F 60 G 38
F 0
G 1 H 35
H 1 I 35
3
A 2 B 10 C 40
B 1 C 20
0
Sample Output
216
30 题意:经典最小生成树。
思路:有向边无向边对生成树来说不重要。顺序读入即可。这里有个小技巧,scanf里的空格可以吃掉连续一串(可以为空)空白符(空格,tab,换行),直到出现第一个非空白符,比getchar()更适用。
#include<stdio.h>
#include<algorithm>
using namespace std; int f[],a[];
struct Edge{
int u,v,w;
}edge[]; bool cmp(Edge a,Edge b)
{
return a.w<b.w;
} int find(int x)
{
return f[x]==x?x:f[x]=find(f[x]);
} int kru(int c,int n)
{
int i;
for(i=;i<=n;i++){
f[i]=i;
}
sort(edge+,edge+c+,cmp);
int cnt=,ans=;
for(i=;i<=c;i++){
int u=edge[i].u;
int v=edge[i].v;
int w=edge[i].w;
int fu=find(u),fv=find(v);
if(fu!=fv){
ans+=w;
f[fv]=fu;
cnt++;
}
if(cnt==n-) break;
}
if(cnt<n-) return -;
else return ans;
}
int main()
{
int n,x,t,c,i,j;
char ch1,ch2;
while(scanf("%d",&n)&&n!=){
c=;
for(i=;i<n;i++){
scanf(" %c %d",&ch1,&t); //
for(j=;j<=t;j++){
scanf(" %c %d",&ch2,&x); //getchar()个别出现RE情况
edge[++c].u=ch1-'A'+;
edge[c].v=ch2-'A'+;
edge[c].w=x;
}
}
printf("%d\n",kru(c,n));
}
return ;
}
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