Constructing Roads 复习了最小生成树的两种算法-蒲公英云

Constructing Roads 复习了最小生成树的两种算法

男娘i 2024-02-18 22:51 147阅读 0赞

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28385 Accepted Submission(s): 10808

Problem Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

3 0 990 692 990 0 179 692 179 0 1 1 2

Sample Output

179

Source

kicc

这条路不好修啊！几乎修了一下午了，

普利姆算法：

他给出了已修好的路，这个就让已通路的村庄的代价为0，就好了，

#include
#include
#include
#include
using namespace std;
int map[101][101],low[101],visit[101];
int Pri(int n)
{
int i,j,sum=0,pos;
for(i=1;i<=n;i++)
{
low[i]=map[1][i];
}
visit[1]=1;
for(i=1;ilow[j])
{

min=low[j];
pos=j;
//cout<<j<<” “<<min<<endl;
}
}

sum+=min;
visit[pos]=1;
for(j=1;j<=n;j++)
if(!visit[j]&&low[j]>map[pos][j])
low[j]=map[pos][j];
}
return sum;
}
int main()
{
int n,i,j,t,x,y;
while(~scanf(“%d”,&n))
{
memset(visit,0,sizeof(visit));
memset(map,0x3f3f3f3f,sizeof(map));
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
cin>>map[i][j];
}
cin>>t;
while(t—)
{
cin>>x>>y;
map[x][y]=0;
map[y][x]=0;
}
cout<<Pri(n)<<endl;
}
return 0;
}

克鲁斯卡尔算法：

#include
#include
#include
#include
using namespace std;
int a[101];
struct kk
{
int s,e,w;
};
kk s[10005];
int cmp(kk x,kk y)
{
return x.w<y.w;
}
int find(int x)
{
if(x==a[x])
return x;
else
return a[x]=find(a[x]);
}

int main()
{
int n;
int i,j,k=0,t,f=0,sum=0,m;
int fx,fy;
//freopen(“1.txt”,”r”,stdin);
while(~scanf(“%d”,&n))
{
k=0;
for(i=1; i<=n; i++)
for(j=1; j<=n; j++)
{
cin>>s[k].w;
s[k].s=i;
s[k].e=j;
k++;
}
sort(s,s+k,cmp);
for(i=1; i<=n; i++)
{
a[i]=i;

}
cin>>t;
for(i=0; i>m>>j;
fx=find(m); 已通路问题：谨慎起见，先让他查找一下，若不相等，则随意让一个连通分量等于另一个连通分量或是查找的结果
fy=find(j);
if(fx!=fy)
a[fx]=a[fy];
}
sum=0;
for(i=0; i0)
{
fx=find(s[i].s);
fy=find(s[i].e);
if(fx!=fy)
{
sum+=s[i].w;
a[fx]=a[fy];