hdoj 2122 Ice_cream’s world III 【最小生成树】

Ice_cream’s world III

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1254 Accepted Submission(s): 414



[align=left]Problem Description[/align]
ice_cream’s world becomes stronger and stronger; every road is built as undirected. The queen enjoys traveling around her world; the queen’s requirement is like II problem, beautifies the roads, by which there are some ways from every
city to the capital. The project’s cost should be as less as better.

[align=left]Input[/align]
Every case have two integers N and M (N<=1000, M<=10000) meaning N cities and M roads, the cities numbered 0…N-1, following N lines, each line contain three integers S, T and C, meaning S connected with T have a road will cost C.

[align=left]Output[/align]
If Wiskey can’t satisfy the queen’s requirement, you must be output “impossible”, otherwise, print the minimum cost in this project. After every case print one blank.

[align=left]Sample Input[/align]

2 1
0 1 10

4 0

[align=left]Sample Output[/align]

10

impossible

代码：

Kruskal算法：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct record
{
int s,e,w;
}num[10010];
bool cmp(record a,record b)
{
return a.w<b.w;
}
int per[1010];
int m,n;
int init()
{
for(int i=0;i<n;i++)
{
per[i]=i;
}
}
int find(int x)
{
int r;
r=x;
while(r!=per[r])
{
r=per[r];
}
per[x]=r;
return r;
}
int join(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
{
per[fx]=fy;
return true;
}
return false;
}
int main()
{
int i,f,sum;
while(scanf("%d%d",&n,&m)!=EOF)
{
f=1;
sum=0;
init();
for(i=0;i<m;i++)
{
scanf("%d%d%d",&num[i].s,&num[i].e,&num[i].w);
}
sort(num,num+m,cmp);
for(i=0;i<m;i++)
{
if(join(num[i].s,num[i].e))
{
sum+=num[i].w;
f++;
}
}
if(f!=n)
printf("impossible\n\n");
else
printf("%d\n\n",sum);
}
return 0;
}

Prim算法：

#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f
int map[1010][1010],low[1010],vis[1010];
int n,m;
int init()
{
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(i==j)
map[i][j]=map[j][i]=0;
else
map[i][j]=map[j][i]=INF;
}
}
}
void prim()
{
int i,j,min,next,sum=0;
for(i=0;i<n;i++)
{
vis[i]=0;
low[i]=map[0][i];
}
vis[0]=1;
for(i=1;i<n;i++)
{
min=INF;
for(j=0;j<n;j++)
{
if(!vis[j]&&min>low[j])
{
min=low[j];
next=j;
}
}
if(min==INF)
{
printf("impossible\n\n");
return ;
}
vis[next]=1;
sum+=min;
for(j=0;j<n;j++)
{
if(!vis[j]&&low[j]>map[next][j])
{
low[j]=map[next][j];
}
}
}
printf("%d\n\n",sum);
}
int main()
{
int i,j,a,b,c;
while(scanf("%d%d",&n,&m)!=EOF)
{
init();
for(i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
if(map[a][b]>c)
map[a][b]=map[b][a]=c;
}
prim();
}
return 0;
}