题意
给定一个无向图（N<=10000, E<=500000），定义f[s,t]表示从s到t经过的每条路径中最长的边的最小值。Q个询问，每个询问一个t，问有多少对(s, t)使得f[s, t] >= t（(s,t)和(t,s)是两对）
思路
按边权从小到大排序。考虑从小到大往图中加边同时计算以加入的边为f值的点对数。
难点和重点在于用并查集维护边的连通情况。
对于新加入的边(u, v)，如果u,v原来便连通，则没有以该边为f值的点对，因为它一定不是最小边。
而如果(u, v)不连通，就可以把这条边能到达的点分为和u连通以及和v连通（分别和u、v在同一并查集中），那么分别从u、 v的连通集中取一个点组成的点对的f值都是这条边，即答案 = 2 * |Union(u)| * |Union(v)|。所以并查集还需要维护各自集中的点个数。
最后用树状数组维护一下前缀和，询问的时候二分+查询前缀和即可。
代码
[cpp]
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#define MID(x,y) ((x+y)/2)
#define MEM(a,b) memset(a,b,sizeof(a))
#define REP(i, begin, end) for (int i = begin; i <= end; i ++)
using namespace std;
const int MAX = 500005;
struct BIT{
int t[MAX<<1];
int bound;
inline void init(int n){
bound = n;
MEM(t, 0);
}
//lowbit(x)：求2^q, q是x二进制最右边的1的位置.
inline int lowbit(int x){
return x & (-x);
}
inline void update(int x, int v){
for (int i = x; i <= bound; i += lowbit(i))
t[i] += v;
}
inline int sum(int x){
int res = 0;
for (int i = x; i >= 1; i -= lowbit(i))
res += t[i];
return res;
}
}bit;
const int MAXN = 10005;
struct Disjoint_Sets{
struct Sets{
int father, ranks, num;
}S[MAXN];
inline void Init(int n){
for (int i = 0; i <= n; i ++){
S[i].father = i;
S[i].ranks = 0;
S[i].num = 1;
}
}
inline int Father(int x){
if (S[x].father == x){
return x;
}
else{
S[x].father = Father(S[x].father); //Path compression
return S[x].father;
}
}
inline bool Union(int x, int y){
int fx = Father(x), fy = Father(y);
if (fx == fy){
return false;
}
else{ //Rank merge
int num1 = S[fx].num, num2 = S[fy].num;
if (S[fx].ranks > S[fy].ranks){
S[fy].father = fx;
}
else{
S[fx].father = fy;
if (S[fx].ranks == S[fy].ranks){
++ S[fy].ranks;
}
}
S[Father(fx)].num = num1+num2;
return true;
}
}
}DS;
struct que{
int t, res;
}q[100005];
struct edges{
int u, v, w;
}e[500005];
bool cmp(edges n1, edges n2){
return n1.w < n2.w;
}
vector <int> adj[10005];
int main(){
int n, m;
while(scanf("%d %d", &n, &m) != EOF){
for (int i = 0; i < m; i ++) scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
sort(e, e+m, cmp);
int Q;
scanf("%d", &Q);
for (int i = 0; i < Q; i ++){
scanf("%d", &q[i].t);
q[i].res = 0;
}
for (int i = 0; i <= n; i ++) adj[i].clear();
DS.Init(n);
bit.init(m);
for (int i = 0; i < m; i ++){
if (DS.Father(e[i].u) != DS.Father(e[i].v)){
bit.update(m - i, 2*DS.S[DS.Father(e[i].u)].num*DS.S[DS.Father(e[i].v)].num);
DS.Union(e[i].u, e[i].v);
}
}
for (int i = 0; i < Q; i ++){
int t = q[i].t;
int l = 0, r = m - 1;
while(l < r){
int mid = MID(l, r);
if (e[mid].w < t){
l = mid + 1;
}
else{
r = mid;
}
}
while (l < m && e[l].w < t) l ++;
printf("%d\n", bit.sum(m - l));
}
}
return 0;
}
[/cpp]