UVa 11857 - Driving Range

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题目

设计电动汽车的电池容量，已知有n个城市，每个城市都有充电站，m条双向道路（可能两个城市间有多条道路），求最小的电池容量。

分析

最小生成树。利用kruskal算法，记录最后加入树的边长即可；如果并查集中的节点数量不是城市数量，则有的城市不连通，输出IMPOSIBALE。

说明

注意数据大小，防止RE。。。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
 
using namespace std;

const int data_size = 1000001;
 
typedef struct _enode
{
	int point1;
	int point2;
	int weight;
}enode;
enode edge[data_size];
 
//union_set
int union_sets[data_size];
int union_rank[data_size];
int union_size[data_size];
 
void union_initial(int a, int b)
{
	for (int i = a; i <= b; ++ i) {
		union_sets[i] = i;
		union_rank[i] = 0;
		union_size[i] = 1;
	}
}
 
int union_find(int a)
{
	if (a != union_sets[a]) {
		union_sets[a] = union_find(union_sets[a]);
	}
	return union_sets[a];
}
 
void union_merge(int a, int b)
{
	if (union_rank[a] < union_rank[b]) {
		union_sets[a] = b;
		union_size[b] += union_size[a];
	} else {
		if (union_rank[a] == union_rank[b]) {
			union_rank[a] ++;
		}
		union_sets[b] = a;
		union_size[a] += union_size[b];
	}
}
//end_union_set

int cmp_e(enode a, enode b)
{
	return a.weight < b.weight;
}
 
void kruskal(int n, int m)
{
	if (m) {
		sort(edge, edge+m, cmp_e);
	}
	
	union_initial(0, n-1);
	int ans = 0;
	for (int i = 0; i < m; ++ i) {
		int A = union_find(edge[i].point1);
		int B = union_find(edge[i].point2);
		if (A != B) { 
			union_merge(A, B);
			ans = edge[i].weight;
		}
	}
	if (union_size[union_find(0)] != n) {
		puts("IMPOSSIBLE");
	} else {
		printf("%d\n", ans);
	}
}
 
int main()
{
	int n, m, a, b, c; 
	while (scanf("%d%d", &n, &m) && n+m) {
		for (int i = 0; i < m; ++ i) {
			scanf("%d%d%d", &a, &b, &c);
			edge[i].point1 = a;
			edge[i].point2 = b;
			edge[i].weight = c;
		}
		
		kruskal(n, m);
	}
	return 0;
}