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枚举+kruskal, 每次更新最小值即可,一次AC,估计把样例过了就能ac,AC代码见下

Posted by su03121231 at 2013-07-29 15:27:53 on Problem 3522
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<limits.h>
typedef struct
{
	int x;
	int y;
	int w;
}init;
int cmp(const void *a, const void *b)
{
	init *c=(init *)a;
	init *d=(init *)b;
	return c->w - d->w;
}
int p[105];
void set()
{
	for(int i=0;i<105;i++)
		p[i]=i;
}
int find(int x)
{
	int s=x;
	while(s!=p[x])
	{
		s=p[x];
		x=p[s];
	}
	return p[x];
}
int main()
{
	int n,m,i,j,min,flag,t,count;
	init map[5005];
	while(scanf("%d%d",&n,&m),n+m)
	{
		memset(map,0,sizeof(map));
		min=INT_MAX;
		flag=0;
		
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&map[i].x,&map[i].y,&map[i].w);
		}
		qsort(map,5005,sizeof(map[0]),cmp);
		for(i=0;!map[i].w;i++);
		for(t=0;t<m-n+2;t++)
		{
			set();
			for(j=i,count=0;j<5005;j++)
			{
				int f1=find(map[j].x);
				int f2=find(map[j].y);
				if(f1 != f2)
				{
					p[f1]=f2;
					count++;
				}
				if(count == n-1)
				{
					flag=1;
					if(map[j].w - map[i].w <min)
						min=map[j].w - map[i].w;
					i++;
					break;
				}
			}
		}
		if(flag)
			printf("%d\n",min);
		else
			printf("-1\n");
	}
	return 0;
}

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