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

Ice_cream’s world III

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1074    Accepted Submission(s): 345

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
   
   
   
   

kruskal：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
	int s,t,c;
}p[1001000];
int cmp(node a,node b)
{
	return a.c < b.c;
}
int n,m;
int per[1100];
void init()
{
	for(int i=0; i < 1010; i++)
		per[i] = i;
}
int find(int x)
{
	if( x == per[x] )
		return x;
	return per[x] = find(per[x]);		
}
int join(int x,int y)
{
	int fx = find(x);
	int fy = find(y);
	if(fx != fy)
	{
		per[fx] = fy;
		return 1;
	}
	return 0;	
}
int main()
{
	int i,j,k;
	while(scanf("%d%d",&n, &m)!=EOF)
	{
		init();
		for(i = 0; i < m; i++)
		{
			scanf("%d%d%d",&p[i].s ,&p[i].t ,&p[i].c );
		}
		sort( p, p+m, cmp);
		int cost = 0;
		for(i = 0; i < m; i++)
		{
			if(join(p[i].s,p[i].t))
				cost += p[i].c;
		}
		int flag = 0;
		for(i = 0; i < n;i++)
		{
			if(per[i] == i)
			{
				flag++;
				if(flag > 1)	break;
			}
		} 
		if(flag > 1)
		puts("impossible\n");
		else
		printf("%d\n\n",cost);
	}
	return 0;
}

prim：

#include<cstdio>
#include<cstring>
#include<cmath>
#define mem(a, b) memset(a, (b), sizeof(a))
#define Wi(a) while(a--)
#define Si(a,b) scanf("%d%d", &a,&b)
#define Pi(a) printf("%d\n\n", (a))
#define INF 0x3f3f3f
int map[1010][1010];
int n,m;
int vis[1010];
int d[1010];
void prim()
{
	mem(vis, 0);
	int i, j, k;
	int ans = 0, minn;
	for(i = 0; i < n; i++)
		d[i] = map[1][i];
	vis[1] = 1;	
	for(i = 1; i < n; i++)
	{
		k = 1;
		minn = INF;
		for(j = 0; j < n; j++)
		{
			if(!vis[j] && d[j] < minn)
			{
				minn = d[j];
				k = j;
			}
		}
		if(minn == INF)//无法联通所有
		{
			puts("impossible\n");return ;
		}
		vis[k] = 1;
		ans += minn;
		for(j = 0; j < n; j++)
		{
			if(!vis[j] && d[j] > map[j][k])
				d[j] = map[j][k];
		}
	}
	Pi(ans);	
}
int main(){
	while(Si(n,m)==2)
	{
		int i,j,k;
		for(i = 0; i < n; i++)
		{
			for(j = 0; j < n; j++)
			{
				map[i][j] = INF;
			}
			map[i][i] = 0;
		}
		int a, b, c;
		Wi(m)
		{
			scanf("%d%d%d", &a, &b, &c); 
			if(map[a][b] > c)
				map[a][b] = map[b][a] = c;
		}
		prim();
	}
	return 0;
}