【NOIP2010提高组】关押罪犯

本题稍难，
排序+并查集即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct pid{int x,y,z;}a[100001];
int f[20001],l[20001][1234];
bool by[20001];
int ans=2147483647,n,m;

int cmp(pid x,pid y) {return x.z>y.z;}

void cz(int x)
{
	by[x]=0;
	for (int i=1;i<=l[x][0];i++)
		if (!by[l[x][i]])
			f[l[x][i]]=f[x]%2+1,cz(l[x][i]);
}

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
		scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
	sort(a+1,a+m+1,cmp);
	for (int i=1;i<=m;i++)
	{
		if (f[a[i].x]!=0&&f[a[i].y]!=0)
		{
			if (f[a[i].x]==f[a[i].y])
			{
				memset(by,0,sizeof(by));
				f[a[i].x]=f[a[i].x]%2+1;cz(a[i].x);
				if (by[a[i].y]) return 0&printf("%d
",a[i].z);
			}
		}
		else if (f[a[i].x]!=0) f[a[i].y]=f[a[i].x]%2+1;
		else if (f[a[i].y]!=0) f[a[i].x]=f[a[i].y]%2+1;
		else f[a[i].x]=1,f[a[i].y]=2;
		l[a[i].x][++l[a[i].x][0]]=a[i].y;
		l[a[i].y][++l[a[i].y][0]]=a[i].x;
	}
	printf("0
");
	return 0;
}