hdu 3790 最短路径模版

给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。

Input

输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点。n和m为0时输入结束。 
(1<n<=1000, 0<m<100000, s != t)

Output

输出 一行有两个数， 最短距离及其花费。

Sample Input

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

Sample Output

9 11

/*
         _...---.._
       ,'          ~~"--..
      /                   ~"-._
     /                         ~-.
    /              .              `-.
    \             -.\                `-.
     \              ~-.                 `-.
  ,-~~\                ~.
 /     \                 `.
.       \                  `.
|        \                   .
|         \                   \
 .         `.                  \
             \                  \
  `           `.                 \
   `           \.                 \
    `           \`.                \
     .           \ -.               \
     `               -.              \
      .           `    -              \  .
      `            \    ~-             \
       `            .     ~.            \
        .            \      -_           \
        `                     -           \
         .            |        ~.          \
         `            |          \          \
          .           |           \          \
          `           |            `.         \
           `          `              \         \
            .          .              `.        `.
            `          :                \         `.
             \         `                 \          `.
              \         .                 `.         `~~-.
               \        :                   `         \   \
                .        .                   \         : `.\
                `        :                    \        |  | .
                 \        .                    \       |  |
                  \       :                     \      `  |  `
                   .                             .      | |_  .
                   `       `.                    `      ` | ~.;
                    \       `.                    .      . .
                     .       `.                   `      ` `
                     `.       `._.                 \      `.\
                      `        <  \                 `.     | .
                       `       `   :                 `     | |
                        `       \                     `    | |
                         `.     |   \                  :  .' |
"Are you crying? "        `     |    \                 `_-'  |
  "It's only the rain."  :    | |   |                   :    ;
"The rain already stopped."`    ; |~-.|                 :    '
  "Devils never cry."       :   \ |                     `   ,
                            `    \`                      :  '
                             :    \`                     `_/
                             `     .\       "For we have none. Our enemy shall fall."
                              `    ` \      "As we apprise. To claim our fate."
                               \    | :     "Now and forever. "
                                \  .'  :    "We'll be together."
                                 :    :    "In love and in hate"
                                 |    .'
                                 |    :     "They will see. We'll fight until eternity. "
                                 |    '     "Come with me.We'll stand and fight together."
                                 |   /      "Through our strength We'll make a better day. "
                                 `_.'       "Tomorrow we shall never surrender."
     sao xin*/
#include <vector>
#include <iostream>
#include <string>
//#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <cstdio>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include<functional>
#define INF 0xfffffff
#define eps 1e-6
#define LL long long

using namespace std;
const int maxn=1e3+5;
const int maxx=1e6+5;
//const double q = (1 + sqrt(5.0)) / 2.0;   // 黄金分割数
/*
std::hex    <<16进制   cin>>std::hex>>a>>std::hex>>b
cout.setf(ios::uppercase);cout<<std::hex<<c<<endl;
b=b>>1; 除以2 二进制运算

//f[i]=(i-1)*(f[i-1]+f[i-2]);   错排
/ for (int i=1; i<=N; i++)
        for (int j=M; j>=1; j--)
        {
            if (weight[i]<=j)
            {
                f[j]=max(f[j],f[j-weight[i]]+value[i]);
            }
        }
priority_queue<int,vector<int>,greater<int> >que3;//注意“>>”会被认为错误，
priority_queue<int,vector<int>,less<int> >que4;////最大值优先
//str
//tmp
//vis
//val
//cnt     2486   hdu 3790最短路径
*/
int n,m;
int map[1005][1005];
int cost[1005][1005];
void dijkstra(int st,int ed)
{
    int i,j,v,Min;
    int visit[1005],dis[1005],value[1005];
    for(i=1;i<=n;i++)
    {
        dis[i]=map[st][i];
        value[i]=cost[st][i];
    }
    memset(visit,0,sizeof(visit));
    visit[st]=1;
    for(i=1;i<n;i++)
    {
        Min=INF;
        for(j=1;j<=n;j++)
        if(!visit[j]&&dis[j]<Min)
        {
            v=j;
            Min=dis[j];
        }
        visit[v]=1;
        for(j=1;j<=n;j++)
        {
            if(!visit[j]&&map[v][j]<INF)
            {
                if(dis[j]>dis[v]+map[v][j])
                {
                    dis[j]=dis[v]+map[v][j];
                    value[j]=value[v]+cost[v][j];
                }
                else if(dis[j]==dis[v]+map[v][j])
                {
                    if(value[j]>value[v]+cost[v][j])
                    value[j]=value[v]+cost[v][j];
                }
            }
        }
    }
    printf("%d %d\n",dis[ed],value[ed]);
}
int main()
{
    int i,j,st,ed;
    int a,b,c,d;
    while(scanf("%d%d",&n,&m),n||m)
    {
        for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
        {
            map[i][j]=INF;
            cost[i][j]=INF;
        }
        while(m--)
        {
            scanf("%d%d%d%d",&a,&b,&c,&d);
            if(map[a][b]>c)
            {
                map[a][b]=map[b][a]=c;
                cost[a][b]=cost[b][a]=d;
            }
            else if(map[a][b]==c)
            {
                if(cost[a][b]>d)
                cost[a][b]=cost[b][a]=d;
            }
        }
        scanf("%d%d",&st,&ed);
        dijkstra(st,ed);
    }
    return 0;
}