Out of Hay
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 18472 | Accepted: 7318 |
Description
The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay situation. There are N (2 <= N <= 2,000) farms (numbered 1..N); Bessie starts at Farm 1. She'll traverse some or all of the M (1 <= M <= 10,000) two-way roads whose length does not exceed 1,000,000,000 that connect the farms. Some farms may be multiply connected with different length roads. All farms are connected one way or another to Farm 1.
Bessie is trying to decide how large a waterskin she will need. She knows that she needs one ounce of water for each unit of length of a road. Since she can get more water at each farm, she's only concerned about the length of the longest road. Of course, she plans her route between farms such that she minimizes the amount of water she must carry.
Help Bessie know the largest amount of water she will ever have to carry: what is the length of longest road she'll have to travel between any two farms, presuming she chooses routes that minimize that number? This means, of course, that she might backtrack over a road in order to minimize the length of the longest road she'll have to traverse.
Bessie is trying to decide how large a waterskin she will need. She knows that she needs one ounce of water for each unit of length of a road. Since she can get more water at each farm, she's only concerned about the length of the longest road. Of course, she plans her route between farms such that she minimizes the amount of water she must carry.
Help Bessie know the largest amount of water she will ever have to carry: what is the length of longest road she'll have to travel between any two farms, presuming she chooses routes that minimize that number? This means, of course, that she might backtrack over a road in order to minimize the length of the longest road she'll have to traverse.
Input
* Line 1: Two space-separated integers, N and M.
* Lines 2..1+M: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, describing a road from A_i to B_i of length L_i.
* Lines 2..1+M: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, describing a road from A_i to B_i of length L_i.
Output
* Line 1: A single integer that is the length of the longest road required to be traversed.
Sample Input
3 3 1 2 23 2 3 1000 1 3 43
Sample Output
43
Hint
OUTPUT DETAILS:
In order to reach farm 2, Bessie travels along a road of length 23. To reach farm 3, Bessie travels along a road of length 43. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a road.
In order to reach farm 2, Bessie travels along a road of length 23. To reach farm 3, Bessie travels along a road of length 43. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a road.
题目描述
Bessie 计划调查N (2 <= N <= 2,000)个农场的干草情况，它从1号农场出发。农场之间总共有M (1 <= M <= 10,000)条双向道路，所有道路的总长度不超过1,000,000,000。有些农场之间存在着多条道路，所有的农场之间都是连通的。
Bessie希望计算出该图中最小生成树中的最长边的长度。
思路
带并查集的最小生成树然后选择不加边而是将每一条加入的边和最小值比较
1 #include<cstdio> 2 #include <iostream> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 struct note 7 { 8 int u; 9 int v; 10 int w; 11 } q[10010]; 12 int f[2010]; 13 int cmp(note a, note b) 14 { 15 return a.w<b.w; 16 } 17 int find(int v) 18 { 19 if (f[v] == v)return v; 20 else return find(f[v]); 21 } 22 int merge(int x, int y) 23 { 24 int t1, t2; 25 t1 = find(x); 26 t2 = find(y); 27 if (t1 != t2) 28 { 29 f[t2] = t1; 30 return 1; 31 } 32 return 0; 33 } 34 int main() 35 { 36 int i, k, n, m; 37 while (~scanf("%d%d", &n, &m)) 38 { 39 for (i = 0; i<m; i++) 40 scanf("%d%d%d", &q[i].u, &q[i].v, &q[i].w); 41 sort(q, q + m, cmp); //按照权值从小到大排序 42 for (i = 1; i <= n; i++) 43 f[i] = i; 44 int count = 0; 45 for (i = 0; i<m; i++) 46 { 47 if (merge(q[i].u, q[i].v))//并查集看是否被连通 48 { 49 count++; 50 if (count == n - 1) 51 { 52 k = i; //最长的一条路的下标 53 break; 54 } 55 } 56 } 57 printf("%d ", q[k].w); 58 } 59 }