Aizu 2249 Road Construction

King Mercer is the king of ACM kingdom. There are one capital and some cities in his kingdom. Amazingly, there are no roads in the kingdom now. Recently, he planned to construct roads between the capital and the cities, but it turned out that the construction cost of his plan is much higher than expected.

In order to reduce the cost, he has decided to create a new construction plan by removing some roads from the original plan. However, he believes that a new plan should satisfy the following conditions:

• For every pair of cities, there is a route (a set of roads) connecting them.
• The minimum distance between the capital and each city does not change from his original plan.

Many plans may meet the conditions above, but King Mercer wants to know the plan with minimum cost. Your task is to write a program which reads his original plan and calculates the cost of a new plan with the minimum cost.

Input

The input consists of several datasets. Each dataset is formatted as follows.

N M

u1 v1 d1 c1 

.

.

.

uM vM dM cM 

The first line of each dataset begins with two integers, N and M (1 ≤ N ≤ 10000, 0 ≤ M ≤ 20000). N and Mindicate the number of cities and the number of roads in the original plan, respectively.

The following M lines describe the road information in the original plan. The i-th line contains four integers, ui, vi, di and ci (1 ≤ ui, vi ≤ N , ui ≠ vi , 1 ≤ di ≤ 1000, 1 ≤ ci ≤ 1000). ui , vi, di and ci indicate that there is a road which connects ui-th city and vi-th city, whose length is di and whose cost needed for construction is ci.

Each road is bidirectional. No two roads connect the same pair of cities. The 1-st city is the capital in the kingdom.

The end of the input is indicated by a line containing two zeros separated by a space. You should not process the line as a dataset.

Output

For each dataset, print the minimum cost of a plan which satisfies the conditions in a line.

Sample Input

3 3
1 2 1 2
2 3 2 1
3 1 3 2
5 5
1 2 2 2
2 3 1 1
1 4 1 1
4 5 1 1
5 3 1 1
5 10
1 2 32 10
1 3 43 43
1 4 12 52
1 5 84 23
2 3 58 42
2 4 86 99
2 5 57 83
3 4 11 32
3 5 75 21
4 5 23 43
5 10
1 2 1 53
1 3 1 65
1 4 1 24
1 5 1 76
2 3 1 19
2 4 1 46
2 5 1 25
3 4 1 13
3 5 1 65
4 5 1 34
0 0
      

Output for the Sample Input

3
5
137
218      

題目要求首都和其他城市的最短距離不變，先跑一下Dijkstra算法求出首都到其他城市的最短距離。接下來求首都到每個城市的最小花費，在保證距離最小時計算花費，求出最小花費即可。      

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <utility>
#include <vector>
using namespace std;
const int maxn =10000+10;
#define INF 0x7fffffff

struct edge{
    int to,d,cost;
};
typedef pair<int,int> P;

int N,M;
vector<edge> G[maxn];
int d[maxn];

void dijkstra(int s){
    priority_queue<P,vector<P>,greater<P> > que;
    fill(d,d+N,INF);
    d[s]=0;
    que.push(P(0,s));

    while(!que.empty()){
        P p=que.top();
        que.pop();

        int v=p.second;
        if(d[v]<p.first) continue;

        for(int i=0;i<G[v].size();i++){
            edge e=G[v][i];
            if(d[e.to]>d[v]+e.d){
                d[e.to]=d[v]+e.d;
                que.push(P(d[e.to],e.to));
            }
        }
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);

    while(cin>>N>>M&&N){
        for(int i=0;i<N;i++){
            G[i].clear();
        }

        for(int i=0;i<M;i++){
            int u,v,x,cost;
            cin>>u>>v>>x>>cost;
            u--;
            v--;
            G[u].push_back((edge){v,x,cost});
            G[v].push_back((edge){u,x,cost});
        }

        dijkstra(0);

        int res=0;
        for(int i=1;i<N;i++){
            int tmp=INF;
            for(int j=0;j<G[i].size();j++){
                if(d[G[i][j].to]+G[i][j].d==d[i]){
                    tmp=min(tmp,G[i][j].cost);
                }
            }
            res+=tmp;
        }
        cout<<res<<endl;
    }
    return 0;
}