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;
}