zoj 3946 Highway Project(优先队列优化dijkstra算法)

Highway Project

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities from the capital as fast as possible. Thus, he proposed the highway project.

The Marjar Empire has N cities (including the capital), indexed from 0 to N - 1 (the capital is 0) and there are M highways can be built. Building the i-th highway costs C_i dollars. It takes D_i minutes to travel between city X_iand Y_i on the i-th highway.

Edward wants to find a construction plan with minimal total time needed to reach other cities from the capital, i.e. the sum of minimal time needed to travel from the capital to city i (1 ≤ i ≤ N). Among all feasible plans, Edward wants to select the plan with minimal cost. Please help him to finish this task.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first contains two integers N, M (1 ≤ N, M ≤ 10⁵).

Then followed by M lines, each line contains four integers X_i, Y_i, D_i, C_i (0 ≤ X_i, Y_i < N, 0 < D_i, C_i < 10⁵).

Output

For each test case, output two integers indicating the minimal total time and the minimal cost for the highway project when the total time is minimized.

Sample Input

2
4 5
0 3 1 1
0 1 1 1
0 2 10 10
2 1 1 1
2 3 1 2
4 5
0 3 1 1
0 1 1 1
0 2 10 10
2 1 2 1
2 3 1 2

Sample Output

4 3
4 4

数量级均为100000，用邻接矩阵会超空间，用结构体存图可能会超时，所以用vector和priority_queue来优化dijkstra即可

#include <stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const long long INF=1e15;
struct node{
	int s;
	int e;
   	long long time;
	long long cost;
};
struct node1{
    long long time;
    long long   cost;
};
struct node2{
	int to;
	long long time;
	long long cost;
	friend bool operator <(node2 a,node2 b)
	{
		if(a.time!=b.time)
		{
			return a.time>b.time;
		}
		return a.cost<b.cost;
	}
};
priority_queue<node2>q;
vector<node>map;
vector<int>f[100005];
node1 minload[100005];
bool visit[100005]; 
int n,m;
void inint()
{
	int i;
	for(i=0;i<=n;i++)
	{
		map.clear();
		f[i].clear();
		minload[i].time=INF,minload[i].cost=INF;
		visit[i]=0;
	}
	while(!q.empty())
	{
		q.pop();
	}
}
int main(int argc, char *argv[])
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        int i,j;
        inint();
        int cnt=0;
        for(i=0;i<m;i++)
        {        
        	int s,e;
			long long time,cost; 
			node have;
            scanf("%d %d %lld %lld",&have.s,&have.e,&have.time,&have.cost);
			map.push_back(have); 
			f[map[map.size()-1].s].push_back(map.size()-1);
           
		    node have1;
			have1.e=have.s,have1.s=have.e,have1.time=have.time,have1.cost=have.cost;
			map.push_back(have1);
            f[map[map.size()-1].s].push_back(map.size()-1);
        }    
        minload[0].cost=0;
        minload[0].time=0;
        node2 go;
		go.to=0;
		go.time=0;
		go.cost=0;    
		q.push(go);
        while(!q.empty())
        {
            node2 next=q.top();
		    q.pop();
            if(visit[next.to])
            continue;	
            visit[next.to]=1;
            for(j=0;j<f[next.to].size();j++)
            {
            	 
                if(minload[map[f[next.to][j]].e].time>map[f[next.to][j]].time+minload[next.to].time)
                {
                    minload[map[f[next.to][j]].e].time=map[f[next.to][j]].time+minload[next.to].time;    
                    minload[map[f[next.to][j]].e].cost=map[f[next.to][j]].cost;
                    
                    node2 add;
                    add.to=map[f[next.to][j]].e;
                    add.time=minload[map[f[next.to][j]].e].time;
                    add.cost=minload[map[f[next.to][j]].e].cost;
                    q.push(add);
                }
               	else  if(minload[map[f[next.to][j]].e].time==map[f[next.to][j]].time+minload[next.to].time&&minload[map[f[next.to][j]].e].cost>map[f[next.to][j]].cost)
                {
                   	minload[map[f[next.to][j]].e].cost=map[f[next.to][j]].cost;
              	 	node2 add;
              	 	add.to=map[f[next.to][j]].e;
                    add.time=minload[map[f[next.to][j]].e].time;
                    add.cost=minload[map[f[next.to][j]].e].cost;
                    q.push(add);
                }
            }
        }
        long long max1time=0;
        long long max1cost=0;
        for(i=1;i<n;i++)
        {
            max1time+=minload[i].time;
            max1cost+=minload[i].cost;
        }
       printf("%lld %lld\n",max1time,max1cost);
    }    
    return 0;
}