aoj2249（最短路）

Description

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
u₁v₁d₁c₁
.
.
.
u_Mv_Md_Mc_M

The first line of each dataset begins with two integers, N and M (1 ≤ N ≤ 10000, 0 ≤ M ≤ 20000). N and M indicate 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, u_i, v_i, d_i and c_i (1 ≤ u_i, v_i≤ N , u_i ≠ v_i , 1 ≤ d_i ≤ 1000, 1 ≤ c_i ≤ 1000). u_i , v_i, d_i and c_i indicate that there is a road which connects u_i-th city and v_i-th city, whose length is d_i and whose cost needed for construction is c_i.

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 31 2 1 22 3 2 13 1 3 25 51 2 2 22 3 1 11 4 1 14 5 1 15 3 1 15 101 2 32 101 3 43 431 4 12 521 5 84 232 3 58 422 4 86 992 5 57 833 4 11 323 5 75 214 5 23 435 101 2 1 531 3 1 651 4 1 241 5 1 762 3 1 192 4 1 462 5 1 253 4 1 133 5 1 654 5 1 340 0

Output for the Sample Input

35137218

在最短路的基础上求最小花费。不过spfa的过程里面有个令人纠结的问题是如何去计算最小花费，因为如果是模拟计算最短路的方式再开一个数组记录最小花费的话，其实是不对的，因为那是计算的从1到每个点的最短花费，并不是整个工程的最小花费，会有重复部分。所以无比纠结。。后来想想，从点1到各个点的最短路一定会形成一个树，所以每个点其实只有一个点指向它，所以我开了一个jl（记录）数组来记录每个点被指向的直接花费，其实比较好理解。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>using namespace std;typedef long long ll;const int maxn=10010,INF=99999999;int ip,n,m,head[maxn],dis[maxn],jl[maxn];bool vis[maxn];struct data{    int next,to,d,w;} tu[maxn*maxn];void init(){    ip=0;    memset(head,-1,sizeof(head));}void add(int a,int b,int d,int w){    tu[ip].to=b,tu[ip].d=d,tu[ip].w=w,tu[ip].next=head[a],head[a]=ip++;}void spfa(){    for(int i=1; i<=n; i++)        dis[i]=jl[i]=INF;    dis[1]=0;    jl[1]=0;    memset(vis,0,sizeof(vis));    queue<int>q;    q.push(1);    while(!q.empty())    {        int t=q.front();        q.pop();        vis[t]=0;        for(int k=head[t]; k!=-1; k=tu[k].next)        {            int d=tu[k].d,v=tu[k].to,w=tu[k].w;            if(dis[t]+d<dis[v])            {                jl[v]=w;///记录最小花费                dis[v]=dis[t]+d;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }            else if(dis[t]+d==dis[v]&&w<jl[v])///记录最小花费                jl[v]=w;        }    }    int ans=0;    for(int i=1; i<=n; i++)        ans+=jl[i];    printf("%d\n",ans);}int main(){    while(~scanf("%d%d",&n,&m)&&m+n)    {        init();        while(m--)        {            int a,b,d,w;            scanf("%d%d%d%d",&a,&b,&d,&w);            add(a,b,d,w);            add(b,a,d,w);        }        spfa();    }    return 0;}