Long time ago, there was a great kingdom and it was being ruled by The Great Arya and Pari The Great. These two had some problems about the numbers they like, so they decided to divide the great kingdom between themselves.
The great kingdom consisted of n cities numbered from 1 to n and m bidirectional roads between these cities, numbered from 1 to m. The i-th road had length equal to wi. The Great Arya and Pari The Great were discussing about destructing some prefix (all road with numbers less than some x) and suffix (all roads with numbers greater than some x) of the roads so there will remain only the roads with numbers l, l + 1, ..., r - 1 and r.
After that they will divide the great kingdom into two pieces (with each city belonging to exactly one piece) such that the hardness of the division is minimized. The hardness of a division is the maximum length of a road such that its both endpoints are in the same piece of the kingdom. In case there is no such road, the hardness of the division is considered to be equal to - 1.
Historians found the map of the great kingdom, and they have q guesses about the l and r chosen by those great rulers. Given these data, for each guess li and ri print the minimum possible hardness of the division of the kingdom.
Input
The first line of the input contains three integers n, m and q (1 ≤ n, q ≤ 1000, 
) — the number of cities and roads in the great kingdom, and the number of guesses, respectively.
The i-th line of the following m lines contains three integers ui, vi and wi (1 ≤ ui, vi ≤ n, 0 ≤ wi ≤ 109), denoting the road number i connects cities ui and vi and its length is equal wi. It's guaranteed that no road connects the city to itself and no pair of cities is connected by more than one road.
Each of the next q lines contains a pair of integers li and ri (1 ≤ li ≤ ri ≤ m) — a guess from the historians about the remaining roads in the kingdom.
Output
For each guess print the minimum possible hardness of the division in described scenario.
Example
5 6 5
5 4 86
5 1 0
1 3 38
2 1 33
2 4 28
2 3 40
3 5
2 6
1 3
2 3
1 6
-1
33
-1
-1
33
题意：给定N点，M无向边。Q次询问，每次询问给出区间[L，R]，现在需要把点集分为两个集合，求最小化集合内的最大边权。
思路：对于这个区间[L,R]，从大到小处理，那么就是求二分图，如果假如长度为X的边染色失败，则答案就是X，因为是一条一条的加边，用2-sat不方便，我们用并查集来解决二分图判定。
（不过整体复杂度还是有些玄学）
1<=x<=N表示一色，N+1<=x<=N+N表示二色，4320ms：
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
struct in{
int u,v,w,id;
bool operator <(const in &x) const { return w>x.w;}
};
in a[maxn*maxn]; int fa[maxn<<],p[maxn*maxn];
int find(int x){
if(x!=fa[x]) fa[x]=find(fa[x]); return fa[x];
}
int un(int x,int y){ int fx=find(x),fy=find(y); fa[fx]=fy; }
int main()
{
int N,M,Q,L,R;
scanf("%d%d%d",&N,&M,&Q);
rep(i,,M) scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].w),a[i].id=i;
sort(a+,a+M+);
rep(i,,Q){
int ans=-; scanf("%d%d",&L,&R);
rep(j,,N+N) fa[j]=j;
rep(j,,M){
if(a[j].id<L||a[j].id>R) continue;
int f1=find(a[j].u),f2=find(a[j].v);
if(f1==f2){ ans=a[j].w; break; }
else un(a[j].u+N,a[j].v),un(a[j].u,a[j].v+N);
}
printf("%d\n",ans);
}
return ;
}
用带权并查集优化，点全部在1<=x<=N范围内，所以理论上复杂度会降低一半，dis表示点到根的距离奇偶性。2932ms。
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
struct in{
int u,v,w,id;
bool operator <(const in &x) const { return w>x.w;}
};
in a[maxn*maxn]; int fa[maxn<<],dis[maxn];
int find(int x){
if(x!=fa[x]){
int fx=find(fa[x]);
dis[x]^=dis[fa[x]];
fa[x]=fx;
}
return fa[x];
}
bool Union(int x,int y){
int fx=find(x),fy=find(y);
if(fx==fy){
if(dis[x]==dis[y]) return false;
return true;
}
fa[fx]=fy; dis[fx]=dis[x]^dis[y]^;
return true;
}
int main()
{
int N,M,Q,L,R;
scanf("%d%d%d",&N,&M,&Q);
rep(i,,M) scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].w),a[i].id=i;
sort(a+,a+M+);
rep(i,,Q){
int ans=-; scanf("%d%d",&L,&R);
rep(j,,N) fa[j]=j,dis[j]=;
rep(j,,M){
if(a[j].id<L||a[j].id>R) continue;
if(!Union(a[j].u,a[j].v)) { ans=a[j].w; break; }
}
printf("%d\n",ans);
}
return ;
}