Aizu - 2249 Road Construction

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<string>
#include<algorithm>
#define MAXSIZE 1000005
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;

//给出若干个建筑之间的一些路，每条路都有对应的长度和需要的花费，
//问在保证源点1 到其他个点的距离最短的情况下，最少的花费是多少

int n,m,vis[MAXSIZE],a[MAXSIZE],k,dist[MAXSIZE],id[MAXSIZE];

struct node
{
    int u;
    int v;
    int w;
    int c;
    int next;
}G[MAXSIZE];

void Add(int u,int v,int w,int c)
{
    G[k].u = u;
    G[k].v = v;
    G[k].w = w;
    G[k].c = c;
    G[k].next = a[u];
    a[u] = k++;
}

int dij()
{
    memset(vis,0,sizeof(vis));
    memset(id,-1,sizeof(id));
    dist[1] = 0;
    int minn ,p;
    for(int i=1;i<k;i++)
    {
        minn = INF;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && dist[j] < minn)
            {
                minn = dist[j];
                p = j;
            }
        }

        if(minn == INF)
            break;
        vis[p] = 1;
        for(int j=a[p];j!=-1;j=G[j].next)
        {
            int v = G[j].v;
            if(dist[v] > dist[p] + G[j].w && !vis[v]) //距离更短就更新
            {
                dist[v] = dist[p] + G[j].w;
                id[v] = j;
            }
            
            //距离相同，花费更小就更新
            else if(id[v]!=-1 && dist[v] == dist[p] + G[j].w && G[j].c < G[id[v]].c && !vis[v])
            {
                id[v] = j;
            }
        }
    }

    int sum=0;
    for(int i=1;i<=n;i++)
    {
        if(id[i] == -1)
            continue;
        else
            sum += G[id[i]].c;
    }
    return sum;
}

void Init()
{
    memset(a,-1,sizeof(a));
    memset(vis,0,sizeof(vis));
    for(int i=0;i<MAXSIZE;i++)
        dist[i] = INF;
    k = 1;
}

int main()
{
    int u,v,w,c;
    while(scanf("%d%d",&n,&m),n+m)
    {
        Init();

        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d%d",&u,&v,&w,&c);
            Add(u,v,w,c);
            Add(v,u,w,c);
        }

        int ans = dij();

        printf("%d\n",ans);
    }
    return 0;
}