POJ 1861 Network (Kruskal求MST模板题)
Time Limit: 1000MS | Memory Limit: 30000K | |||
Total Submissions: 14103 | Accepted: 5528 | Special Judge |
Description
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem — not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.
Input
Output
Sample Input
4 6
1 2 1
1 3 1
1 4 2
2 3 1
3 4 1
2 4 1
Sample Output
1
4
1 2
1 3
2 3
3 4
Source
题目链接:poj.org/problem?id=1861
题目大意:n个点,m条线,每条线有个权值,现在要求最长的路最短且让各个点都连通,求最短的最长路,边个数和对应边
题目分析:样例有问题,应该是
1
4
1 2
1 3
3 4
裸的Kruskal注意这里要求最长路最短,而Kruskal正好是对权值从小到大排序后的贪心算法
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; int const MAX = 15005; int fa[MAX]; int n, m, ma, num; int re1[MAX], re2[MAX]; struct Edge { int u, v, w; }e[MAX]; bool cmp(Edge a, Edge b) { return a.w < b.w; } void UF_set() { for(int i = 0; i < MAX; i++) fa[i] = i; } int Find(int x) { return x == fa[x] ? x : fa[x] = Find(fa[x]); } void Union(int a, int b) { int r1 = Find(a); int r2 = Find(b); if(r1 != r2) fa[r2] = r1; } void Kruskal() { UF_set(); for(int i = 0; i < m; i++) { int u = e[i].u; int v = e[i].v; if(Find(u) != Find(v)) { re1[num] = u; re2[num] = v; Union(u, v); ma = max(ma, e[i].w); num ++; } if(num >= n - 1) break; } } int main() { ma = 0; num = 0; scanf("%d %d", &n, &m); 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); Kruskal(); printf("%d\n%d\n", ma, num); for(int i = 0; i < num; i++) printf("%d %d\n", re1[i], re2[i]); }
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