HDU 3790 簡單最短路徑問題（dijkstra+雙重權值）

給你n個點，m條無向邊，每條邊都有長度d和花費p，給你起點s終點t，要求輸出起點到終點的最短距離及其花費，如果最短距離有多條路線，則輸出花費最少的。

Input輸入n,m，點的編號是1~n,然後是m行，每行4個數 a,b,d,p，表示a和b之間有一條邊，且其長度爲d，花費爲p。最後一行是兩個數 s,t;起點s，終點。n和m爲0時輸入結束。 
(1<n<=1000, 0<m<100000, s != t) Output輸出 一行有兩個數， 最短距離及其花費。 Sample Input

3 2
1 2 5 6
2 3 4 5
1 3
0 0

Sample Output

9 11

思路：這也是這道題的重點，去掉重複的邊，如果邊相同，選擇金錢比較少的邊

上代碼

#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define Max 1010
int n, pp[Max][Max], M[Max][Max], d[Max], cost[Max];
bool visit[Max];
vector<pair<int, int> >G[Max];
void init(int nn)
{
	for (int i = 1; i <= nn; i++)
	{
		for (int j = 1; j <= nn; j++)
		{
			M[i][j] = inf;
			pp[i][j] = inf;
		}
		d[i] = inf;
		cost[i] = inf;
	}
	memset(visit, false, sizeof(visit));
}
void dijkstra(int s, int t)
{
	int minv;
	d[s] = 0, cost[s] = 0;
	while (1)
	{
		minv = inf;
		int u = -1;
		for (int i = 1; i <= n; i++)
		{
			if (minv>d[i] && visit[i] == false)
			{
				u = i;
				minv = d[i];
			}
		}
		if (u == -1)
			break;
		visit[u] = true;
		for (int i = 1; i <= n; i++)
		{
			if (visit[i])
				continue;
			if (d[i] == d[u] + M[u][i] && M[u][i] != inf&&cost[i]>cost[u] + pp[u][i])
				cost[i] = cost[u] + pp[u][i];
			if (d[i]>d[u] + M[u][i])
			{
				d[i] = d[u] + M[u][i];
				cost[i] = cost[u] + pp[u][i];
			}
		}
	}
	cout << d[t] << " " << cost[t] << endl;
}
int main()
{
	//freopen("Text1.txt","r",stdin);
	int m, u, v, d, p, s, t;
	while (scanf("%d%d", &n, &m) != EOF)
	{
		if (n == 0 && m == 0)
			break;
		init(n);
		for (int i = 0; i<m; i++)
		{
			scanf("%d%d%d%d", &u, &v, &d, &p);
			if (M[u][v]>d)
			{
				M[u][v] = M[v][u] = d;
				pp[u][v] = pp[v][u] = p;
			}
			if (M[u][v] == d&&pp[u][v] > p)//邊相同選擇最小金錢的邊，否則去掉
				pp[u][v] = pp[v][u] = p;
		}
		scanf("%d%d", &s, &t);
		dijkstra(s, t);
	}
	return 0;
}