Sicily 1090. Highways 解题报告

题目链接：Sicily 1090

思路：

简单的最小生成树问题，这里用prim算法即可。用visited数组记录每个结点是否已经被访问，即是否已经在最小生成树中。每次从不在最小生成树中的结点中取出一个key值最小的结点放入生成树中，key值表示结点到已经在生成树中点集合的最小距离。每次加入一个结点后更新与它相邻的结点的key值。

代码：

#include <iostream>
#include <queue>
#include <stdio.h>
#include <memory.h>
using namespace std;

const int MAX = 501;
const int MAX_LEN = 65537;

int prim_get_longest(int n, int adj[][MAX]);

int main() {
	int testcases;
	cin >> testcases;
	while(testcases--){
		int n;
		cin >> n;
		int adj[MAX][MAX];
		for (int i = 1; i <= n; ++i) {
			for (int j = 1; j <= n; ++j) {
				cin >> adj[i][j];
			}
		}
		cout << prim_get_longest(n, adj) << endl;
		if(testcases > 0)
			cout << endl;
	}
	return 0;
}

int prim_get_longest(int n, int adj[][MAX]){
	//Post: Using Prim algorithm to find the minimal spanning tree,
	// 		and return the longest edge in the tree.
	int longest = 0;
	bool visited[MAX];
	int key[MAX];
	memset(visited, false, sizeof(visited));
	for (int i = 2; i <= n; ++i) {
		key[i] = adj[1][i];
	}
	visited[1] = true; //root
	for (int i = 1; i < n; ++i) {
		int u, min = MAX_LEN;
		for (int j = 2; j <= n; ++j) {
			if(!visited[j] && key[j] < min){
				min = key[j];
				u = j;
			}
		}
		longest = max(longest, min);
		visited[u] = true;
		for (int v = 2; v <= n; ++v) {
			if(!visited[v] && adj[v][u] < key[v]){
				key[v] = adj[v][u];
			}
		}
	}
	return longest;
}

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Sicily 1090. Highways 解题报告

原文：http://www.cnblogs.com/jolin123/p/3915471.html