hdu 3938 Portal(离线并查集)
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Portal
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 747 Accepted Submission(s): 373
Problem Description
ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of portals will cost min{T} energies. T in a path between point V and point U is the length of the longest edge in the path. There may be lots of paths between two points. Now ZLGG owned L energies and he want to know how many kind of path he could make.
Input
There are multiple test cases. The first line of input contains three integer N, M and Q (1 < N ≤ 10,000, 0 < M ≤ 50,000, 0 < Q ≤ 10,000). N is the number of points, M is the number of edges and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, 0 ≤ c ≤ 10^8) describing an edge connecting the point a and b with cost c. Each of the following Q lines contain a single integer L (0 ≤ L ≤ 10^8).
Output
Output the answer to each query on a separate line.
Sample Input
10 10 107 2 16 8 34 5 85 8 22 8 96 4 52 1 58 10 57 3 77 8 810615918276
Sample Output
36131133613621613题意:给出m条边,每条边都有一个花费,每两点a到b可形成多条路径,定义走过a b路径的花费为a到b的各条路径中花费最大的边的最小值,给出q个询问,每个询问给出拥有多少权值,对于每个询问,要输出能走过的路径条数。思路:类似kruskal算法的思想,用并查集离线处理。将互达的点看成一个集合,集合的点的个数为rank[i]。按边的花费从小到大给边排序,然后从小到大枚举每条边,若这条边的两个端点不在一个同集合中,设分别为集合a和集合b,那么这条边必为集合a的点到集合b的点的路径中花费最大的边的最小值,那么答案就可以加上rank[a]*rank[b]。AC代码:#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#define L(r) (r<<1)#define R(r) (r<<1|1)using namespace std;const int maxn = 10005;const int maxm = 50005;const int INF = 1000000000;int n, m, q;int fa[maxn], rank[maxn];struct Edge{ int u, v, w;} et[maxm];struct node{ int L, id;} query[maxn];bool cmp1(Edge a, Edge b){ return a.w < b.w;}bool cmp2(node a, node b){ return a.L < b.L;}void init(){ for(int i = 1; i <= n; i++) { rank[i] = 1; fa[i] = i; }}int find(int x){ return fa[x] == x ? x : fa[x] = find(fa[x]);}void Union(int a, int b){ if(rank[a] >= rank[b]) { fa[b] = a; rank[a] += rank[b]; } else { fa[a] = b; rank[b] += rank[a]; }}int main(){ while(~scanf("%d%d%d", &n, &m, &q)) { init(); for(int i = 0; i < m; i++) scanf("%d%d%d", &et[i].u, &et[i].v, &et[i].w); for(int i = 0; i < q; i++) { scanf("%d", &query[i].L); query[i].id = i; } sort(et, et + m, cmp1); sort(query, query + q, cmp2); int cnt = 0; int ans[maxm] = {0}; for(int i = 0; i < q; i++) { if(i > 0) ans[query[i].id] = ans[query[i - 1].id]; while(et[cnt].w <= query[i].L && cnt < m) { int ra = find(et[cnt].u); int rb = find(et[cnt].v); if(ra != rb) { ans[query[i].id] += rank[ra] * rank[rb]; Union(ra, rb); } cnt++; } } for(int i = 0; i < q; i++) printf("%d\n", ans[i]); } return 0;}
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