hdu3371 Connect the Cities (MST)
Connect the Cities
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13722 Accepted Submission(s): 3711
want to take too much money.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected
cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6
1
pid=1102" target="_blank" style="color:rgb(26,92,200); text-decoration:none">1102
1301 1162pid=1198" target="_blank" style="color:rgb(26,92,200); text-decoration:none">1198
1598Statistic | Submit | Discuss | Note
难理解的就是最后那k行。開始的数字t表示有几个城市。然后输入t个城市,表示第一个城市和第二个连接,第二个和第三个连接。
。。
用kruskal算法超时的多提交两次。
。当然也能够用pri算法。。
不想写。。
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
struct node
{
int a,b,cost;
}c[30000];
int fa[505];
void init(int n)
{
for(int i=1;i<=n;i++)
fa[i]=i;
}
bool cmp(node x,node y)
{
return x.cost<y.cost;
}
int find(int x)
{
if(fa[x]!=x) fa[x]=find(fa[x]);
return fa[x];
}
int main()
{
int n,k,m,ncase;
scanf("%d",&ncase);
while(ncase--)
{
scanf("%d %d %d",&n,&k,&m);
init(n);
for(int i=0;i<k;i++)
scanf("%d %d %d",&c[i].a,&c[i].b,&c[i].cost);
for(int i=1;i<=m;i++)
{
int x,pos,pos1;
scanf("%d %d",&x,&pos);
for(int j=1;j<x;j++)
{
scanf("%d",&pos1);
c[k].a=pos,c[k].b=pos1,c[k].cost=0;
pos=pos1;
k++;
}
}
sort(c,c+k,cmp);
int sum=0;
for(int i=0;i<k;i++)
{
int x=find(c[i].a);
int y=find(c[i].b);
if(x!=y)
sum+=c[i].cost,fa[x]=y;
}
int count=0;
for(int i=1;i<=n;i++)
if(fa[i]==i)
count++;
if(count!=1)
printf("-1\n");
else
printf("%d\n",sum);
}
return 0;
}
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