Poj 2395 最小生成树(Kruskal 算法)

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Out of Hay
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 13359 Accepted: 5153
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.
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.
    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.

想法:Bessie要走过每一个村庄,而且这连接的道路要最短,因为每一个村庄都可以供水,所以要求出其中最长的一条边。于是我们就想到了最小生成树的方法。

#include <iostream>#include <sstream>#include <ios>#include <iomanip>#include <functional>#include <algorithm>#include <vector>#include <string>#include <list>#include <queue>#include <deque>#include <stack>#include <set>#include <map>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <climits>#include <cctype>using namespace std;int par[1100];int r[1100];void init(int n){    for (int i = 0; i<n; i++) {        par[i] = i;        r[i] = 0;    }}int find(int x){    if (par[x] == x) {        return x;    }    return par[x] = find(par[x]);}void unite(int x,int y){    x = find(x);    y = find(y);    if (x == y) {        return;    }    if (r[x]<r[y]) {        par[x] = y;    }    else    {        par[y] = x;        if (r[x] == r[y]) {            r[x]++;        }    }}bool same(int x,int y){    return find(x) == find(y);}struct edge{    int u,v,cost;};bool comp(const edge& e1,const edge& e2){    return e1.cost < e2.cost;}edge es[11000];int n,m;int kruskal(){    sort(es, es+m, comp);    init(n);    int res = 0;    for (int i = 0; i<m; i++) {        edge e = es[i];        if (!same(e.u, e.v)) {            unite(e.u, e.v);            res = max(res,e.cost);            //用res记录最长边。        }    }    return res;}int main(){    while (cin>>n>>m) {        for (int i = 0; i<m; i++) {            int a,b,c;            cin>>a>>b>>c;            a--;b--;            es[i].u = a;            es[i].v = b;            es[i].cost = c;        }        int sum = kruskal();        cout<<sum<<endl;    }    return 0;}
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