九度 题目1008：最短路径问题

dijkstra求最短路，出现相同最短路时取花费最小的那条。

代码：

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>

using namespace std;

int n, m;
int map[1010][1010];
bool vi[1010];
int cost[1010][1010];
int dis[1010];
int price[1010];
int s, t;

const int INF = 2100000000;

void init()
{
	for (int i = 1; i <= n; ++ i)
	{
		vi[i] = false;
		dis[i] = INF;
		price[i] = INF;
		for (int j = 1; j <= n; ++ j)
		{
			map[i][j] = INF;
		}
	}
	for (int i = 0; i < m; ++ i)
	{
		int from, to, d, p;
		scanf("%d%d%d%d", &from, &to, &d, &p);
		map[to][from] = map[from][to] = d;
		cost[to][from] = cost[from][to] = p;
	}
	scanf("%d%d", &s, &t);	
}

void dijkstra()
{
	dis[s] = 0;
	price[s] = 0;		
	for (int k = 0; k < n; ++ k)
	{
		int index = -1;
		int minn = INF;
		for (int i = 1; i <= n; ++ i)
		{
			if (vi[i]==false && dis[i]<minn)
			{
				index = i;
				minn = dis[i];
			}
		}
		vi[index] = true;		
		for (int j = 1; j <= n; ++ j)
		{
			if (dis[j] > dis[index] + map[index][j])
			{
				dis[j] = dis[index] + map[index][j];
				price[j] = price[index] + cost[index][j]; 
			} else if (dis[j] == dis[index] + map[index][j])
			{
				price[j] = min(price[j], price[index]+cost[index][j]);
			}
		}
	}
}

int main()
{
	while (cin >> n >> m, n || m)
	{
		init();
		dijkstra();
		cout << dis[t] << " " << price[t] << endl;
	}
	return 0;
}