Ice_cream’s world III

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

Total Submission(s): 926    Accepted Submission(s): 303

Problem Description

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.

Input

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.

Output

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.

Sample Input

2 1
0 1 10

4 0

Sample Output

10

impossible

Author

Wiskey

Source

HDU 2007-10 Programming Contest_WarmUp

最小生成树：prime 算法，还有克鲁斯卡尔算法。

两者都可以处理。

prime代码如下：

#include<stdio.h>
#include<string.h>
int v[10010],map[10010][10010];
#define inf 0xfffffff;
int main()
{
	int n,m,i,j,a,b,c;
	while(~scanf("%d%d",&n,&m))
	{
		int sum=0;
		memset(v,0,sizeof(v));
		for(i=0;i<n;i++)//初始化
		{
			for(j=0;j<n;j++)
			{
				map[i][j]=inf;
			}
		}
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			if(c<map[a][b])//判断是否是重边，要是起点终点都相同的时候，就去权值最小的作为两个顶点间的距离即是权值
			map[a][b]=map[b][a]=c;
		}
		v[0]=1;
		int flag;
		for(i=1;i<n;i++)
		{
			int min=inf;
			flag=-1;
			for(j=0;j<n;j++)
			{
				if(!v[j]&&map[0][j]<min)
				{
					min=map[0][j];
					flag=j;
				}
			}
			if(flag==-1)
			break;
			v[flag]=1;
			sum+=map[0][flag];
			for(j=0;j<n;j++)
			{
				if(!v[j]&&map[0][j]>map[flag][j])
				map[0][j]=map[flag][j];
			}
		}
		if(flag==-1)
		printf("impossible\n\n");
		else
		printf("%d\n\n",sum);
	}
	return 0;
}

kuskal算法代码如：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
	int a,b,c;
}s[10010];
int father[10010];
int find(int r)
{
	return r==father[r]?r:find(father[r]);
}
int cmp(node x,node y)
{
	return x.c<y.c;
}
int main()
{
	int n,m,i,j;
	while(~scanf("%d%d",&n,&m))
	{
		for(i=0;i<n;i++)//初始化
		{
			father[i]=i;
		}
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&s[i].a,&s[i].b,&s[i].c);
		}
		sort(s,s+m,cmp);//求的是最小的生成树，所以按照权值 增序排列
		int count=0,sum=0;
		for(i=0;i<m;i++)//一共有m条边，所有循环到m
		{
			int fa=find(s[i].a);
			int fb=find(s[i].b);
			if(fa!=fb)
			{
				father[fa]=fb;
				sum+=s[i].c;
				count++;//除了这一点都是克鲁斯卡尔算法模板
			}
		}
		if(count!=n-1)//是否是n-1条边 即建立了一个满足条件的最小生成树
		{
			printf("impossible\n\n");
		}
		else
		printf("%d\n\n",sum);
	}
	return 0;
}