HDU1102 Constructing Roads，Prim算法

最新推荐文章于 2020-09-04 09:26:37 发布

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#prim #hdu 1102 #Constructing Roads

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本文深入探讨了如何使用Prim算法解决村庄间道路建设问题，旨在寻找连接所有村庄所需的最短总路程。通过具体实例，详细解释了算法步骤，并提供了完整的代码实现。

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Constructing Roads

Problem Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

3
0 990 692
990 0 179
692 179 0
1
1 2

Sample Output

179

题目变态之处在于没有说是多组测试数据，估计每人都贡献了一个wa


#include<iostream>
#include<string>
#include<string.h>
#include<vector>
#include<cstdio>
#include<set>
#include<memory>
#include<stack>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;

int mm[200];
int cost[200][200];
bool used[200];
int inf = 1 << 29;
int n;

int prim()
{
	for (int i = 1; i <= n; i++)
	{
		mm[i] = inf;
		used[i] = false;
	}
	int res = 0;
	mm[1] = 0;
	while (1)
	{
		int v = -1;
		for (int i = 1; i <= n; i++)
			if (!used[i] && (v == -1 || mm[i] < mm[v]))
				v = i;

			if (v == -1) break;
			used[v] = true;
			res += mm[v];
			for (int j = 1; j <= n; j++)
				mm[j] = min(mm[j], cost[v][j]);
	}
	return res;
}

int main()
{
	ios::sync_with_stdio(false);
	while (cin >> n)
	{
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				cin >> cost[i][j];
		int m;
		cin >> m;
		int a, b;
		for (int i = 0; i < m; i++)
			cin >> a >> b, cost[a][b] = 0, cost[b][a] = 0;
		cout << prim() << endl;
	}
}