hdu 2066 一个人的旅行 最短路

       这题简单的最短路，一开始我还傻傻地去用bfs写，WA了几次后，发现用错方法了，用bfs求最短路，是对每条边权值都相同的图才可以的，后来改用单源最短路径， 这题数据规模比较大（a,b去到1000）不能用floyd算法。

 

#include <iostream>
#include <cstring>
using namespace std;

const int MAX = 1005;
const int INF = 1000000000;

int arcs[MAX][MAX];
int S,T,D;
int start_place[MAX];
int size;
bool isfound[MAX];
int cost[MAX];
int goal_place[MAX];

int shortest(int v)
{
	memset(isfound, false, sizeof(isfound));

	for (int i = 1; i <= size; i++)
		cost[i] = arcs[v][i];

	isfound[v] = true;

	int index;
	int _min;

	for (int i = 1; i < size; i++)
	{
		_min = INF;
		for (int j = 1; j <= size; j++)
			if (!isfound[j] && _min > cost[j])
				_min = cost[index = j];

		if (_min == INF) break;

		isfound[index] = true;

		for (int j = 1; j <= size; j++)
			if (!isfound[j] && cost[j] > cost[index] + arcs[index][j])
				cost[j] = cost[index] + arcs[index][j];
	}

	_min = INF;
	for (int i = 0; i < D; i++)	
		_min = min(_min, cost[goal_place[i]]);

	return _min;
}
int main()
{
	int a, b, c;
	int num;
	int ans;

	while (cin >> T >> S >> D)
	{

		for (int i = 0; i < MAX; i++)
		{
			for (int j = 0; j < MAX; j++)
				arcs[i][j] = INF;
			arcs[i][i] = 0;
		}

		for (int i = 0; i < T; i++)
		{
			cin >> a >> b >> c;
			arcs[a][b] = min(arcs[a][b], c);
			arcs[b][a] = arcs[a][b];
			size = max(size,max(a,b));
		}

		for (int i = 0; i < S; i++)
			cin >> start_place[i];
		for (int i = 0; i < D; i++)
			cin >> goal_place[i];
		
		ans = INF;
		for (int i = 0; i < S; i++)
			ans = min(ans, shortest(start_place[i]));
		
		cout << ans << endl;
	}
	return 0;
}