题意：一张无向图，q个查询，对于每个x，有多少对点之间的路径中最长的一条路不大于x。
思路：比赛时王秋平写的，我补下题。这题也比较简单，将边排序，从小到大加到并查集，对查询也排序，从小到大对于每个查询把不大于x的边加到并查集，用cnt[y]记录以y为根的连通块有多少节点，那么在连通块发生 变化时，ans=2 * cnt[x] * cnt[y]
1 #include <iostream> 2 #include <cstdio> 3 #include <fstream> 4 #include <algorithm> 5 #include <cmath> 6 #include <deque> 7 #include <vector> 8 #include <queue> 9 #include <string> 10 #include <cstring> 11 #include <map> 12 #include <stack> 13 #include <set> 14 #define LL long long 15 #define eps 1e-8 16 #define INF 0x3f3f3f3f 17 #define MAXN 20005 18 #define MAXM 100005 19 #define MAXP 5005 20 using namespace std; 21 struct Edge{ 22 int from, to, dist; 23 Edge(int from = 0, int to = 0, int dist = 0) :from(from), to(to), dist(dist) {}; 24 }edges[MAXM]; 25 int father[MAXN], cnt[MAXN], res[MAXP]; 26 struct Node{ 27 int x, pos; 28 Node(int x = 0, int pos = 0) :x(x), pos(pos){}; 29 }; 30 Node a[MAXP]; 31 bool compare(Edge a, Edge b){ 32 return a.dist < b.dist; 33 } 34 bool comp(Node a, Node b){ 35 return a.x < b.x; 36 } 37 int find(int x){ 38 if (father[x] == x){ 39 return x; 40 } 41 father[x] = find(father[x]); 42 return father[x]; 43 } 44 int main() 45 { 46 #ifndef ONLINE_JUDGE 47 freopen("in.txt", "r", stdin); 48 //freopen("out.txt", "w", stdout); 49 #endif // OPEN_FILE 50 int T; 51 scanf("%d", &T); 52 int n, m, q; 53 while (T--){ 54 scanf("%d%d%d", &n, &m, &q); 55 int x, y, z; 56 for (int i = 1; i <= m; i++){ 57 scanf("%d%d%d", &x, &y, &z); 58 edges[i] = Edge(x, y, z); 59 } 60 sort(edges + 1, edges + m + 1, compare); 61 for (int i = 1; i <= q; i++){ 62 scanf("%d", &a[i].x); 63 a[i].pos = i; 64 } 65 sort(a + 1, a + q + 1, comp); 66 for (int i = 1; i <= n; i++){ 67 father[i] = i; 68 cnt[i] = 1; 69 } 70 int s = 1; 71 int ans = 0; 72 memset(res, 0, sizeof(res)); 73 for (int i = 1; i <= q; i++){ 74 for (int j = s; j <= m && edges[j].dist <= a[i].x; j++, s = j){ 75 int x = find(edges[j].from); 76 int y = find(edges[j].to); 77 if (x == y) continue; 78 father[x] = y; 79 ans += 2 * cnt[x] * cnt[y]; 80 cnt[y] += cnt[x]; 81 } 82 res[a[i].pos] = ans; 83 } 84 for (int i = 1; i <= q; i++){ 85 printf("%d ", res[i]); 86 } 87 } 88 }